Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $

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$$ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $$

Using the L'Hospital Rule I obtained the value -1/4 but the answer is given to be -1/3. I can't find the Mistake.
Here's what I did, please point out the mistake

calculus limits

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edited 2 days ago

amWhy

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asked Sep 9 at 9:28

Archit Jain

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  Would you show the steps you did?
  – Mark
  Sep 9 at 9:31
  

  
  
  
  


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  Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
  – Masacroso
  Sep 9 at 9:34
  
  

  
  
  
  


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  Okay I'll try those. Thanks
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


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  By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
  – Teepeemm
  2 days ago
  

  
  
  
  


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  @Teepeemm Okay. Thanks that second part is what I really didn't know
  – Archit Jain
  2 days ago
  

  
  
  
  

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show 12 more comments

up vote
5
down vote

favorite

1

$$ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $$

Using the L'Hospital Rule I obtained the value -1/4 but the answer is given to be -1/3. I can't find the Mistake.
Here's what I did, please point out the mistake

calculus limits

share|cite|improve this question

edited 2 days ago

amWhy

190k26221433

asked Sep 9 at 9:28

Archit Jain

495

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Would you show the steps you did?
  – Mark
  Sep 9 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
  – Masacroso
  Sep 9 at 9:34
  
  

  
  
  
  


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  Okay I'll try those. Thanks
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
  – Teepeemm
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Teepeemm Okay. Thanks that second part is what I really didn't know
  – Archit Jain
  2 days ago
  

  
  
  
  

 | 
show 12 more comments

up vote
5
down vote

favorite

1

up vote
5
down vote

favorite

1

1

$$ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $$

Using the L'Hospital Rule I obtained the value -1/4 but the answer is given to be -1/3. I can't find the Mistake.
Here's what I did, please point out the mistake

calculus limits

share|cite|improve this question

edited 2 days ago

amWhy

190k26221433

asked Sep 9 at 9:28

Archit Jain

495

$$ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $$

Using the L'Hospital Rule I obtained the value -1/4 but the answer is given to be -1/3. I can't find the Mistake.
Here's what I did, please point out the mistake

calculus limits

calculus limits

share|cite|improve this question

edited 2 days ago

amWhy

190k26221433

asked Sep 9 at 9:28

Archit Jain

495

share|cite|improve this question

edited 2 days ago

amWhy

190k26221433

asked Sep 9 at 9:28

Archit Jain

495

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

amWhy

190k26221433

edited 2 days ago

amWhy

190k26221433

edited 2 days ago

amWhy

190k26221433

190k26221433

asked Sep 9 at 9:28

Archit Jain

495

asked Sep 9 at 9:28

Archit Jain

495

asked Sep 9 at 9:28

Archit Jain

495

495

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Would you show the steps you did?
  – Mark
  Sep 9 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
  – Masacroso
  Sep 9 at 9:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I'll try those. Thanks
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
  – Teepeemm
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Teepeemm Okay. Thanks that second part is what I really didn't know
  – Archit Jain
  2 days ago
  

  
  
  
  

 | 
show 12 more comments

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  Would you show the steps you did?
  – Mark
  Sep 9 at 9:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
  – Masacroso
  Sep 9 at 9:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay I'll try those. Thanks
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
  – Teepeemm
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Teepeemm Okay. Thanks that second part is what I really didn't know
  – Archit Jain
  2 days ago
  

  
  
  
  

9

9

Would you show the steps you did?
– Mark
Sep 9 at 9:31

Would you show the steps you did?
– Mark
Sep 9 at 9:31

Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
– Masacroso
Sep 9 at 9:34

Using L'Hôpital you need to apply at least twice. In this case is easier to use the asymptotic approximation $sin xsim_0 x$. As is said above: please, share your calculations.
– Masacroso
Sep 9 at 9:34

Okay I'll try those. Thanks
– Archit Jain
Sep 9 at 9:43

Okay I'll try those. Thanks
– Archit Jain
Sep 9 at 9:43

1

1

By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
– Teepeemm
2 days ago

By the way, your mistake is that you can't split the limit into two pieces unless you know the separate limits exist. Having done that, you can't take $sin x+xcos x$ from one piece and put it into the other piece.
– Teepeemm
2 days ago

@Teepeemm Okay. Thanks that second part is what I really didn't know
– Archit Jain
2 days ago

@Teepeemm Okay. Thanks that second part is what I really didn't know
– Archit Jain
2 days ago

 | 
show 12 more comments

9 Answers
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up vote
10
down vote

Hint: Write the function as $$fracsin^2(x)-x^2x^4times fracx^2sin^2(x)$$ Otherwise use the Talor's expantion if you know it.

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answered Sep 9 at 9:37

mrs

58.1k650141

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  And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
  – Mark
  Sep 9 at 9:40
  
  

  
  
  
  


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  I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


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  @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
  – gimusi
  Sep 9 at 9:57
  

  
  
  
  


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  @mrs That's the good method to simplify by standard limits!
  – gimusi
  Sep 9 at 9:58
  

  
  
  
  


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  Yeah I got it now. Thanks
  – Archit Jain
  Sep 9 at 9:58
  

  
  
  
  

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up vote
9
down vote

By l'Hopital we have

$$lim_x to 0frac1x^2 - frac1 sin^2 x =lim_x to 0fracsin^2 x-x^2x^2sin^2 x$$

$$stackrelH.R.=lim_x to 0fracsin 2x-2x2xsin^2 x+x^2sin 2x $$

$$stackrelH.R.=lim_x to 0frac2cos 2x-22sin^2 x+2xsin 2x+2xsin 2x +2x^2cos 2x$$

$$stackrelH.R.=lim_x to 0frac-4sin 2x2sin 2 x+8xcos 2x+4 sin 2x+4xcos 2x-4x^2sin 2x$$

$$stackrelH.R.=lim_x to 0frac-8cos 2x12cos 2 x+8cos 2x-16x sin 2x-8xsin 2x+4cos 2x-8xsin 2x-8x^2cos2x$$

$$=lim_x to 0frac-8cos 2x24cos 2 x-32x sin 2x-8x^2cos2x =frac-824-0-0=-frac13$$

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edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

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  Good steps! Had a hard job in typing. Great++
  – mrs
  Sep 9 at 9:53
  

  
  
  
  


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  definitively you are masochist. +1 for typing this monstrosity :)
  – Masacroso
  Sep 9 at 10:12
  

  
  
  
  


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  @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
  – gimusi
  Sep 9 at 10:14
  

  
  
  
  


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  I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
  – Paramanand Singh
  Sep 9 at 11:45
  

  
  
  
  


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  @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
  – gimusi
  Sep 9 at 12:06
  

  
  
  
  

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up vote
8
down vote

As an alternative by Taylor expansion as $xto 0$

$$sin x = x -frac16x^3 + o(x^3)implies sin^2 x = left(x -frac16x^3 + o(x^3)right)^2=x^2-frac13x^4+o(x^4)$$

we have

$$frac1x^2 - frac1 sin^2 x =fracsin^2 x-x^2x^2sin^2 x=fracx^2-frac13x^4+o(x^4)-x^2x^2left(x^2-frac13x^4+o(x^4)right)=$$$$=frac-frac13x^4+o(x^4)x^4+o(x^4)=frac-frac13+o(1)1+o(1)to -frac13$$

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edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

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  You're missing the square, can't edit it because it's <6 letters
  – Sudix
  Sep 9 at 11:40
  

  
  
  
  


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  @Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
  – gimusi
  Sep 9 at 11:42
  
  

  
  
  
  

add a comment | 

up vote
7
down vote

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)=lim_x to 0fracsin^2 x-x^2x^2sin^2 x=lim_x to 0frac(sin x-x)(sin x+x)x^4$$
$$=lim_x to 0frac(sin x+x)xlim_x to 0fracx(sin x-x)x^4=lim_x to 0frac2x(sin x-x)x^4=lim_x to 0frac2(sin x-x)x^3$$
$$=lim_x to 0frac2(cos x-1)3x^2=lim_x to 0frac-2sin x6x=frac-13.$$

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edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

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  Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
  – Archit Jain
  Sep 9 at 9:45
  

  
  
  
  


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  The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
  – gimusi
  Sep 9 at 9:59
  
  

  
  
  
  


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  @gimusi and the final step also need some justification.
  – Masacroso
  Sep 9 at 10:13
  

  
  
  
  


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  @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
  – gimusi
  Sep 9 at 10:15
  

  
  
  
  


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  just $limlimits_xto 0fracsin x+xx=2$.
  – Riemann
  Sep 9 at 10:22
  

  
  
  
  

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up vote
3
down vote

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $sin(x)$. Using the Taylor expansion:

$$sin(x) = x - fracx^36 + fracx^5120 +mathcalO(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $mathcalO(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $dfrac1sin^2(x)$:

$$frac1sin^2(x) = frac1x^2left[1 - fracx^26 + fracx^4120 +mathcalO(x^6)right]^-2$$

To expand the square brackets, we can use:

$$frac1(1+u)^2 = 1-2 u + 3 u^2 + mathcalO(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - fracx^26 + fracx^4120 +mathcalO(x^6)$. We have:

$$u^2 = left[- fracx^26 + fracx^4120 +mathcalO(x^6)right]^2 = fracx^436 +mathcalO(x^6)$$

Therefore:

$$frac11+u= 1-2 u + 3 u^2 +mathcalO(u^3)= 1 + fracx^23 + fracx^415 +mathcalO(x^6)$$

And we see that:

$$frac1sin^2(x) = frac1x^2 + frac13 + fracx^215 +mathcalO(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $dfrac1sin^4(x)$ by squaring both sides of this expansion, like:

$$lim_xto 0left[frac1sin^4(x)-frac1x^4 - frac23 x^2right]= frac1145$$

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answered Sep 9 at 15:22

Count Iblis

7,96621332

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  @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
  – Count Iblis
  Sep 9 at 17:07
  

  
  
  
  


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  ops yes of course!
  – gimusi
  Sep 9 at 17:22
  

  
  
  
  


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  I've added also a solution by that way!
  – gimusi
  2 days ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$sin x = x-frac16 x^3+o(x^3) implies frac1sin x=frac 1xleft(1-frac16x^2+o(x^2)right)^-1=frac1x+frac16x+o(x)$$

therefore

$$left( frac1x^2 - frac1 sin^2 x right)
=left( frac1x + frac1 sin x right) left( frac1x - frac1 sin x right)=$$
$$=left(frac2x+frac16x+o(x)right) left( -frac16x+o(x)right)
=-frac13+o(1) to -frac13$$

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answered 2 days ago

gimusi

71.4k73786

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up vote
2
down vote

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

---

What you can do instead (notice the asymmetry):

$$lim_xto0fracsin^2x-x^2x^2sin^2x=lim_xto0fracsin^2x-x^2x^4=lim_xto0fracsin x+xxlim_xto0fracsin x-xx^3
\=2lim_xto0fraccos x-13x^2=-2lim_xto0fracsin x6x=-frac13.$$

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edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

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up vote
1
down vote

$$lim_xto0frac(sinx+x)(sinx-x)xsinxcdot xsinx$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$lim_xto0fracsinx-xx^3=-frac16$$

$$lim_xto0fracx-tanxx^3=-frac13$$

$$lim_xto 0frace^x-1-xx^2=frac12$$

So using this,

$$lim_xto0fracx^2sin^2xcdot frac(sinx+x)xcdot frac(sinx-x)x^3$$

$$1cdot2cdot -frac16$$

$$-frac13$$

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answered 2 days ago

prog_SAHIL

1,464318

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up vote
1
down vote

As noticed in the comments, we are allowed to proceed as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)= lim_x to 0 left( fracsin^2 x-x^2x^2sin^2 x right)=lim_x to 0 left( fracsin x+xxsin x right)left( fracsin x-xxsin x right)=ldots$$

but we are not allowed to proceed as follows

$$ldots=lim_x to 0 left( fracsin x+xxsin x right)lim_x to 0left( fracsin x-xxsin x right)=ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$ldots=lim_x to 0 frac cos x+1 sin x+xcos xcdot lim_x to 0 frac cos x-1 sin x+xcos x=ldots$$

and the LHS limit, in the form $frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ ldots=lim_x to 0 (cos x+1),lim_x to 0 frac cos x-1 (sin x+xcos x)^2=ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

---

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)
= lim_x to 0left(fracsin^2 x-x^2x^4cdotfracx^2sin^2 xright)
stackrel? = lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =ldots$$

and since, using l'Hopital for each part, we have

$$lim_x to 0fracsin^2 x-x^2x^4=lim_x to 0fracsin 2x-2x4x^3=lim_x to 0frac2cos 2x-212x^2=lim_x to 0frac-4sin 2x24x=lim_x to 0frac-8cos 2x24=-frac13$$

$$lim_x to 0fracx^2sin^2 x =lim_x to 0frac2xsin 2x =lim_x to 0frac22cos 2x =1$$

we see that the initial step is allowed and then we can conclude that

$$ldots= lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =-frac13cdot 1 =-frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $lim_x to 0fracsin x x=1$.

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edited 5 hours ago

answered yesterday

gimusi

71.4k73786

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up vote
10
down vote

Hint: Write the function as $$fracsin^2(x)-x^2x^4times fracx^2sin^2(x)$$ Otherwise use the Talor's expantion if you know it.

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answered Sep 9 at 9:37

mrs

58.1k650141

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
  – Mark
  Sep 9 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
  – gimusi
  Sep 9 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrs That's the good method to simplify by standard limits!
  – gimusi
  Sep 9 at 9:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I got it now. Thanks
  – Archit Jain
  Sep 9 at 9:58
  

  
  
  
  

 | 
show 1 more comment

up vote
10
down vote

Hint: Write the function as $$fracsin^2(x)-x^2x^4times fracx^2sin^2(x)$$ Otherwise use the Talor's expantion if you know it.

share|cite|improve this answer

answered Sep 9 at 9:37

mrs

58.1k650141

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
  – Mark
  Sep 9 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
  – gimusi
  Sep 9 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrs That's the good method to simplify by standard limits!
  – gimusi
  Sep 9 at 9:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I got it now. Thanks
  – Archit Jain
  Sep 9 at 9:58
  

  
  
  
  

 | 
show 1 more comment

up vote
10
down vote

up vote
10
down vote

Hint: Write the function as $$fracsin^2(x)-x^2x^4times fracx^2sin^2(x)$$ Otherwise use the Talor's expantion if you know it.

share|cite|improve this answer

answered Sep 9 at 9:37

mrs

58.1k650141

Hint: Write the function as $$fracsin^2(x)-x^2x^4times fracx^2sin^2(x)$$ Otherwise use the Talor's expantion if you know it.

share|cite|improve this answer

answered Sep 9 at 9:37

mrs

58.1k650141

share|cite|improve this answer

share|cite|improve this answer

answered Sep 9 at 9:37

mrs

58.1k650141

answered Sep 9 at 9:37

mrs

58.1k650141

answered Sep 9 at 9:37

mrs

58.1k650141

58.1k650141

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
  – Mark
  Sep 9 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
  – gimusi
  Sep 9 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrs That's the good method to simplify by standard limits!
  – gimusi
  Sep 9 at 9:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I got it now. Thanks
  – Archit Jain
  Sep 9 at 9:58
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
  – Mark
  Sep 9 at 9:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
  – Archit Jain
  Sep 9 at 9:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
  – gimusi
  Sep 9 at 9:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mrs That's the good method to simplify by standard limits!
  – gimusi
  Sep 9 at 9:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah I got it now. Thanks
  – Archit Jain
  Sep 9 at 9:58
  

  
  
  
  

1

1

And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
– Mark
Sep 9 at 9:40

And also it will help if you have memorized the "small angle approximation" $x/sin(x) rightarrow 1$ as $x rightarrow 0$
– Mark
Sep 9 at 9:40

I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
– Archit Jain
Sep 9 at 9:43

I couldn't upload my steps right now. So I'll try these too. And yeah I know the binomial as well as the taylor approximations. But how to use L'Hospital on it?
– Archit Jain
Sep 9 at 9:43

@ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
– gimusi
Sep 9 at 9:57

@ArchitJain You can refer to standard limit for the second one and use l'Hopital for the first one, which is simpler.
– gimusi
Sep 9 at 9:57

@mrs That's the good method to simplify by standard limits!
– gimusi
Sep 9 at 9:58

@mrs That's the good method to simplify by standard limits!
– gimusi
Sep 9 at 9:58

Yeah I got it now. Thanks
– Archit Jain
Sep 9 at 9:58

Yeah I got it now. Thanks
– Archit Jain
Sep 9 at 9:58

 | 
show 1 more comment

up vote
9
down vote

By l'Hopital we have

$$lim_x to 0frac1x^2 - frac1 sin^2 x =lim_x to 0fracsin^2 x-x^2x^2sin^2 x$$

$$stackrelH.R.=lim_x to 0fracsin 2x-2x2xsin^2 x+x^2sin 2x $$

$$stackrelH.R.=lim_x to 0frac2cos 2x-22sin^2 x+2xsin 2x+2xsin 2x +2x^2cos 2x$$

$$stackrelH.R.=lim_x to 0frac-4sin 2x2sin 2 x+8xcos 2x+4 sin 2x+4xcos 2x-4x^2sin 2x$$

$$stackrelH.R.=lim_x to 0frac-8cos 2x12cos 2 x+8cos 2x-16x sin 2x-8xsin 2x+4cos 2x-8xsin 2x-8x^2cos2x$$

$$=lim_x to 0frac-8cos 2x24cos 2 x-32x sin 2x-8x^2cos2x =frac-824-0-0=-frac13$$

share|cite|improve this answer

edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Good steps! Had a hard job in typing. Great++
  – mrs
  Sep 9 at 9:53
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  definitively you are masochist. +1 for typing this monstrosity :)
  – Masacroso
  Sep 9 at 10:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
  – gimusi
  Sep 9 at 10:14
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
  – Paramanand Singh
  Sep 9 at 11:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
  – gimusi
  Sep 9 at 12:06
  

  
  
  
  

 | 
show 3 more comments

up vote
9
down vote

By l'Hopital we have

$$lim_x to 0frac1x^2 - frac1 sin^2 x =lim_x to 0fracsin^2 x-x^2x^2sin^2 x$$

$$stackrelH.R.=lim_x to 0fracsin 2x-2x2xsin^2 x+x^2sin 2x $$

$$stackrelH.R.=lim_x to 0frac2cos 2x-22sin^2 x+2xsin 2x+2xsin 2x +2x^2cos 2x$$

$$stackrelH.R.=lim_x to 0frac-4sin 2x2sin 2 x+8xcos 2x+4 sin 2x+4xcos 2x-4x^2sin 2x$$

$$stackrelH.R.=lim_x to 0frac-8cos 2x12cos 2 x+8cos 2x-16x sin 2x-8xsin 2x+4cos 2x-8xsin 2x-8x^2cos2x$$

$$=lim_x to 0frac-8cos 2x24cos 2 x-32x sin 2x-8x^2cos2x =frac-824-0-0=-frac13$$

share|cite|improve this answer

edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Good steps! Had a hard job in typing. Great++
  – mrs
  Sep 9 at 9:53
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  definitively you are masochist. +1 for typing this monstrosity :)
  – Masacroso
  Sep 9 at 10:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
  – gimusi
  Sep 9 at 10:14
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
  – Paramanand Singh
  Sep 9 at 11:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
  – gimusi
  Sep 9 at 12:06
  

  
  
  
  

 | 
show 3 more comments

up vote
9
down vote

up vote
9
down vote

By l'Hopital we have

$$lim_x to 0frac1x^2 - frac1 sin^2 x =lim_x to 0fracsin^2 x-x^2x^2sin^2 x$$

$$stackrelH.R.=lim_x to 0fracsin 2x-2x2xsin^2 x+x^2sin 2x $$

$$stackrelH.R.=lim_x to 0frac2cos 2x-22sin^2 x+2xsin 2x+2xsin 2x +2x^2cos 2x$$

$$stackrelH.R.=lim_x to 0frac-4sin 2x2sin 2 x+8xcos 2x+4 sin 2x+4xcos 2x-4x^2sin 2x$$

$$stackrelH.R.=lim_x to 0frac-8cos 2x12cos 2 x+8cos 2x-16x sin 2x-8xsin 2x+4cos 2x-8xsin 2x-8x^2cos2x$$

$$=lim_x to 0frac-8cos 2x24cos 2 x-32x sin 2x-8x^2cos2x =frac-824-0-0=-frac13$$

share|cite|improve this answer

edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

By l'Hopital we have

$$lim_x to 0frac1x^2 - frac1 sin^2 x =lim_x to 0fracsin^2 x-x^2x^2sin^2 x$$

$$stackrelH.R.=lim_x to 0fracsin 2x-2x2xsin^2 x+x^2sin 2x $$

$$stackrelH.R.=lim_x to 0frac2cos 2x-22sin^2 x+2xsin 2x+2xsin 2x +2x^2cos 2x$$

$$stackrelH.R.=lim_x to 0frac-4sin 2x2sin 2 x+8xcos 2x+4 sin 2x+4xcos 2x-4x^2sin 2x$$

$$stackrelH.R.=lim_x to 0frac-8cos 2x12cos 2 x+8cos 2x-16x sin 2x-8xsin 2x+4cos 2x-8xsin 2x-8x^2cos2x$$

$$=lim_x to 0frac-8cos 2x24cos 2 x-32x sin 2x-8x^2cos2x =frac-824-0-0=-frac13$$

share|cite|improve this answer

edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

share|cite|improve this answer

share|cite|improve this answer

edited Sep 9 at 10:01

edited Sep 9 at 10:01

edited Sep 9 at 10:01

answered Sep 9 at 9:48

gimusi

71.4k73786

answered Sep 9 at 9:48

gimusi

71.4k73786

answered Sep 9 at 9:48

gimusi

71.4k73786

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Good steps! Had a hard job in typing. Great++
  – mrs
  Sep 9 at 9:53
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  definitively you are masochist. +1 for typing this monstrosity :)
  – Masacroso
  Sep 9 at 10:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
  – gimusi
  Sep 9 at 10:14
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
  – Paramanand Singh
  Sep 9 at 11:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
  – gimusi
  Sep 9 at 12:06
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Good steps! Had a hard job in typing. Great++
  – mrs
  Sep 9 at 9:53
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  definitively you are masochist. +1 for typing this monstrosity :)
  – Masacroso
  Sep 9 at 10:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
  – gimusi
  Sep 9 at 10:14
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
  – Paramanand Singh
  Sep 9 at 11:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
  – gimusi
  Sep 9 at 12:06
  

  
  
  
  

1

1

Good steps! Had a hard job in typing. Great++
– mrs
Sep 9 at 9:53

Good steps! Had a hard job in typing. Great++
– mrs
Sep 9 at 9:53

4

4

definitively you are masochist. +1 for typing this monstrosity :)
– Masacroso
Sep 9 at 10:12

definitively you are masochist. +1 for typing this monstrosity :)
– Masacroso
Sep 9 at 10:12

1

1

@Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
– gimusi
Sep 9 at 10:14

@Masacroso I just wanted to convince myself to do never use l'Hopital blindly! :P
– gimusi
Sep 9 at 10:14

5

5

I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
– Paramanand Singh
Sep 9 at 11:45

I hope readers are convinced (via this answer) that one should not apply L'Hospital's Rule blindly!! +1
– Paramanand Singh
Sep 9 at 11:45

1

1

@farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
– gimusi
Sep 9 at 12:06

@farruhota Yes of course. That's precisely the method suggested by mrs in the accepted answer. Bye
– gimusi
Sep 9 at 12:06

 | 
show 3 more comments

up vote
8
down vote

As an alternative by Taylor expansion as $xto 0$

$$sin x = x -frac16x^3 + o(x^3)implies sin^2 x = left(x -frac16x^3 + o(x^3)right)^2=x^2-frac13x^4+o(x^4)$$

we have

$$frac1x^2 - frac1 sin^2 x =fracsin^2 x-x^2x^2sin^2 x=fracx^2-frac13x^4+o(x^4)-x^2x^2left(x^2-frac13x^4+o(x^4)right)=$$$$=frac-frac13x^4+o(x^4)x^4+o(x^4)=frac-frac13+o(1)1+o(1)to -frac13$$

share|cite|improve this answer

edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You're missing the square, can't edit it because it's <6 letters
  – Sudix
  Sep 9 at 11:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
  – gimusi
  Sep 9 at 11:42
  
  

  
  
  
  

add a comment | 

up vote
8
down vote

As an alternative by Taylor expansion as $xto 0$

$$sin x = x -frac16x^3 + o(x^3)implies sin^2 x = left(x -frac16x^3 + o(x^3)right)^2=x^2-frac13x^4+o(x^4)$$

we have

$$frac1x^2 - frac1 sin^2 x =fracsin^2 x-x^2x^2sin^2 x=fracx^2-frac13x^4+o(x^4)-x^2x^2left(x^2-frac13x^4+o(x^4)right)=$$$$=frac-frac13x^4+o(x^4)x^4+o(x^4)=frac-frac13+o(1)1+o(1)to -frac13$$

share|cite|improve this answer

edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You're missing the square, can't edit it because it's <6 letters
  – Sudix
  Sep 9 at 11:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
  – gimusi
  Sep 9 at 11:42
  
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

As an alternative by Taylor expansion as $xto 0$

$$sin x = x -frac16x^3 + o(x^3)implies sin^2 x = left(x -frac16x^3 + o(x^3)right)^2=x^2-frac13x^4+o(x^4)$$

we have

$$frac1x^2 - frac1 sin^2 x =fracsin^2 x-x^2x^2sin^2 x=fracx^2-frac13x^4+o(x^4)-x^2x^2left(x^2-frac13x^4+o(x^4)right)=$$$$=frac-frac13x^4+o(x^4)x^4+o(x^4)=frac-frac13+o(1)1+o(1)to -frac13$$

share|cite|improve this answer

edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

As an alternative by Taylor expansion as $xto 0$

$$sin x = x -frac16x^3 + o(x^3)implies sin^2 x = left(x -frac16x^3 + o(x^3)right)^2=x^2-frac13x^4+o(x^4)$$

we have

$$frac1x^2 - frac1 sin^2 x =fracsin^2 x-x^2x^2sin^2 x=fracx^2-frac13x^4+o(x^4)-x^2x^2left(x^2-frac13x^4+o(x^4)right)=$$$$=frac-frac13x^4+o(x^4)x^4+o(x^4)=frac-frac13+o(1)1+o(1)to -frac13$$

share|cite|improve this answer

edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

share|cite|improve this answer

share|cite|improve this answer

edited Sep 9 at 11:42

edited Sep 9 at 11:42

edited Sep 9 at 11:42

answered Sep 9 at 10:10

gimusi

71.4k73786

answered Sep 9 at 10:10

gimusi

71.4k73786

answered Sep 9 at 10:10

gimusi

71.4k73786

71.4k73786

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You're missing the square, can't edit it because it's <6 letters
  – Sudix
  Sep 9 at 11:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
  – gimusi
  Sep 9 at 11:42
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You're missing the square, can't edit it because it's <6 letters
  – Sudix
  Sep 9 at 11:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
  – gimusi
  Sep 9 at 11:42
  
  

  
  
  
  

1

1

You're missing the square, can't edit it because it's <6 letters
– Sudix
Sep 9 at 11:40

You're missing the square, can't edit it because it's <6 letters
– Sudix
Sep 9 at 11:40

@Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
– gimusi
Sep 9 at 11:42

@Sudix Yes of course! Thanks for pointing that out, I fix it! Bye
– gimusi
Sep 9 at 11:42

add a comment | 

up vote
7
down vote

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)=lim_x to 0fracsin^2 x-x^2x^2sin^2 x=lim_x to 0frac(sin x-x)(sin x+x)x^4$$
$$=lim_x to 0frac(sin x+x)xlim_x to 0fracx(sin x-x)x^4=lim_x to 0frac2x(sin x-x)x^4=lim_x to 0frac2(sin x-x)x^3$$
$$=lim_x to 0frac2(cos x-1)3x^2=lim_x to 0frac-2sin x6x=frac-13.$$

share|cite|improve this answer

edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
  – Archit Jain
  Sep 9 at 9:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
  – gimusi
  Sep 9 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @gimusi and the final step also need some justification.
  – Masacroso
  Sep 9 at 10:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
  – gimusi
  Sep 9 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  just $limlimits_xto 0fracsin x+xx=2$.
  – Riemann
  Sep 9 at 10:22
  

  
  
  
  

 | 
show 4 more comments

up vote
7
down vote

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)=lim_x to 0fracsin^2 x-x^2x^2sin^2 x=lim_x to 0frac(sin x-x)(sin x+x)x^4$$
$$=lim_x to 0frac(sin x+x)xlim_x to 0fracx(sin x-x)x^4=lim_x to 0frac2x(sin x-x)x^4=lim_x to 0frac2(sin x-x)x^3$$
$$=lim_x to 0frac2(cos x-1)3x^2=lim_x to 0frac-2sin x6x=frac-13.$$

share|cite|improve this answer

edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
  – Archit Jain
  Sep 9 at 9:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
  – gimusi
  Sep 9 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @gimusi and the final step also need some justification.
  – Masacroso
  Sep 9 at 10:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
  – gimusi
  Sep 9 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  just $limlimits_xto 0fracsin x+xx=2$.
  – Riemann
  Sep 9 at 10:22
  

  
  
  
  

 | 
show 4 more comments

up vote
7
down vote

up vote
7
down vote

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)=lim_x to 0fracsin^2 x-x^2x^2sin^2 x=lim_x to 0frac(sin x-x)(sin x+x)x^4$$
$$=lim_x to 0frac(sin x+x)xlim_x to 0fracx(sin x-x)x^4=lim_x to 0frac2x(sin x-x)x^4=lim_x to 0frac2(sin x-x)x^3$$
$$=lim_x to 0frac2(cos x-1)3x^2=lim_x to 0frac-2sin x6x=frac-13.$$

share|cite|improve this answer

edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)=lim_x to 0fracsin^2 x-x^2x^2sin^2 x=lim_x to 0frac(sin x-x)(sin x+x)x^4$$
$$=lim_x to 0frac(sin x+x)xlim_x to 0fracx(sin x-x)x^4=lim_x to 0frac2x(sin x-x)x^4=lim_x to 0frac2(sin x-x)x^3$$
$$=lim_x to 0frac2(cos x-1)3x^2=lim_x to 0frac-2sin x6x=frac-13.$$

share|cite|improve this answer

edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

share|cite|improve this answer

share|cite|improve this answer

edited Sep 9 at 12:14

edited Sep 9 at 12:14

edited Sep 9 at 12:14

answered Sep 9 at 9:41

Riemann

3,0481321

answered Sep 9 at 9:41

Riemann

3,0481321

answered Sep 9 at 9:41

Riemann

3,0481321

3,0481321

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
  – Archit Jain
  Sep 9 at 9:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
  – gimusi
  Sep 9 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @gimusi and the final step also need some justification.
  – Masacroso
  Sep 9 at 10:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
  – gimusi
  Sep 9 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  just $limlimits_xto 0fracsin x+xx=2$.
  – Riemann
  Sep 9 at 10:22
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
  – Archit Jain
  Sep 9 at 9:45
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
  – gimusi
  Sep 9 at 9:59
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @gimusi and the final step also need some justification.
  – Masacroso
  Sep 9 at 10:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
  – gimusi
  Sep 9 at 10:15
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  just $limlimits_xto 0fracsin x+xx=2$.
  – Riemann
  Sep 9 at 10:22
  

  
  
  
  

1

1

Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
– Archit Jain
Sep 9 at 9:45

Thank You. I was breaking the limit into two separate product of limits. Maybe I made a mistake doing that
– Archit Jain
Sep 9 at 9:45

2

2

The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
– gimusi
Sep 9 at 9:59

The intermediate step $$ldots =lim_x to 0frac(sin x-x)(sin x+x)x^4$$ $$=lim_x to 0frac2x(sin x-x)x^4=ldots $$ should be justified. Are you using taylor's expansion?
– gimusi
Sep 9 at 9:59

1

1

@gimusi and the final step also need some justification.
– Masacroso
Sep 9 at 10:13

@gimusi and the final step also need some justification.
– Masacroso
Sep 9 at 10:13

@Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
– gimusi
Sep 9 at 10:15

@Masacroso Yes even if it is just a standard limit and I can uderstand that, but the intermediat step can be considere wrong as it is presented.
– gimusi
Sep 9 at 10:15

2

2

just $limlimits_xto 0fracsin x+xx=2$.
– Riemann
Sep 9 at 10:22

just $limlimits_xto 0fracsin x+xx=2$.
– Riemann
Sep 9 at 10:22

 | 
show 4 more comments

up vote
3
down vote

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $sin(x)$. Using the Taylor expansion:

$$sin(x) = x - fracx^36 + fracx^5120 +mathcalO(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $mathcalO(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $dfrac1sin^2(x)$:

$$frac1sin^2(x) = frac1x^2left[1 - fracx^26 + fracx^4120 +mathcalO(x^6)right]^-2$$

To expand the square brackets, we can use:

$$frac1(1+u)^2 = 1-2 u + 3 u^2 + mathcalO(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - fracx^26 + fracx^4120 +mathcalO(x^6)$. We have:

$$u^2 = left[- fracx^26 + fracx^4120 +mathcalO(x^6)right]^2 = fracx^436 +mathcalO(x^6)$$

Therefore:

$$frac11+u= 1-2 u + 3 u^2 +mathcalO(u^3)= 1 + fracx^23 + fracx^415 +mathcalO(x^6)$$

And we see that:

$$frac1sin^2(x) = frac1x^2 + frac13 + fracx^215 +mathcalO(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $dfrac1sin^4(x)$ by squaring both sides of this expansion, like:

$$lim_xto 0left[frac1sin^4(x)-frac1x^4 - frac23 x^2right]= frac1145$$

share|cite|improve this answer

answered Sep 9 at 15:22

Count Iblis

7,96621332

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
  – Count Iblis
  Sep 9 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ops yes of course!
  – gimusi
  Sep 9 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've added also a solution by that way!
  – gimusi
  2 days ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $sin(x)$. Using the Taylor expansion:

$$sin(x) = x - fracx^36 + fracx^5120 +mathcalO(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $mathcalO(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $dfrac1sin^2(x)$:

$$frac1sin^2(x) = frac1x^2left[1 - fracx^26 + fracx^4120 +mathcalO(x^6)right]^-2$$

To expand the square brackets, we can use:

$$frac1(1+u)^2 = 1-2 u + 3 u^2 + mathcalO(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - fracx^26 + fracx^4120 +mathcalO(x^6)$. We have:

$$u^2 = left[- fracx^26 + fracx^4120 +mathcalO(x^6)right]^2 = fracx^436 +mathcalO(x^6)$$

Therefore:

$$frac11+u= 1-2 u + 3 u^2 +mathcalO(u^3)= 1 + fracx^23 + fracx^415 +mathcalO(x^6)$$

And we see that:

$$frac1sin^2(x) = frac1x^2 + frac13 + fracx^215 +mathcalO(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $dfrac1sin^4(x)$ by squaring both sides of this expansion, like:

$$lim_xto 0left[frac1sin^4(x)-frac1x^4 - frac23 x^2right]= frac1145$$

share|cite|improve this answer

answered Sep 9 at 15:22

Count Iblis

7,96621332

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
  – Count Iblis
  Sep 9 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ops yes of course!
  – gimusi
  Sep 9 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've added also a solution by that way!
  – gimusi
  2 days ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $sin(x)$. Using the Taylor expansion:

$$sin(x) = x - fracx^36 + fracx^5120 +mathcalO(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $mathcalO(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $dfrac1sin^2(x)$:

$$frac1sin^2(x) = frac1x^2left[1 - fracx^26 + fracx^4120 +mathcalO(x^6)right]^-2$$

To expand the square brackets, we can use:

$$frac1(1+u)^2 = 1-2 u + 3 u^2 + mathcalO(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - fracx^26 + fracx^4120 +mathcalO(x^6)$. We have:

$$u^2 = left[- fracx^26 + fracx^4120 +mathcalO(x^6)right]^2 = fracx^436 +mathcalO(x^6)$$

Therefore:

$$frac11+u= 1-2 u + 3 u^2 +mathcalO(u^3)= 1 + fracx^23 + fracx^415 +mathcalO(x^6)$$

And we see that:

$$frac1sin^2(x) = frac1x^2 + frac13 + fracx^215 +mathcalO(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $dfrac1sin^4(x)$ by squaring both sides of this expansion, like:

$$lim_xto 0left[frac1sin^4(x)-frac1x^4 - frac23 x^2right]= frac1145$$

share|cite|improve this answer

answered Sep 9 at 15:22

Count Iblis

7,96621332

My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $sin(x)$. Using the Taylor expansion:

$$sin(x) = x - fracx^36 + fracx^5120 +mathcalO(x^7)$$

Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $mathcalO(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $dfrac1sin^2(x)$:

$$frac1sin^2(x) = frac1x^2left[1 - fracx^26 + fracx^4120 +mathcalO(x^6)right]^-2$$

To expand the square brackets, we can use:

$$frac1(1+u)^2 = 1-2 u + 3 u^2 + mathcalO(u^3)$$

This can be derived by differentiating the geometric series term by term. We can then substitute $u = - fracx^26 + fracx^4120 +mathcalO(x^6)$. We have:

$$u^2 = left[- fracx^26 + fracx^4120 +mathcalO(x^6)right]^2 = fracx^436 +mathcalO(x^6)$$

Therefore:

$$frac11+u= 1-2 u + 3 u^2 +mathcalO(u^3)= 1 + fracx^23 + fracx^415 +mathcalO(x^6)$$

And we see that:

$$frac1sin^2(x) = frac1x^2 + frac13 + fracx^215 +mathcalO(x^4)$$

The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $dfrac1sin^4(x)$ by squaring both sides of this expansion, like:

$$lim_xto 0left[frac1sin^4(x)-frac1x^4 - frac23 x^2right]= frac1145$$

share|cite|improve this answer

answered Sep 9 at 15:22

Count Iblis

7,96621332

share|cite|improve this answer

share|cite|improve this answer

answered Sep 9 at 15:22

Count Iblis

7,96621332

answered Sep 9 at 15:22

Count Iblis

7,96621332

answered Sep 9 at 15:22

Count Iblis

7,96621332

7,96621332

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
  – Count Iblis
  Sep 9 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ops yes of course!
  – gimusi
  Sep 9 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've added also a solution by that way!
  – gimusi
  2 days ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
  – Count Iblis
  Sep 9 at 17:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ops yes of course!
  – gimusi
  Sep 9 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I've added also a solution by that way!
  – gimusi
  2 days ago
  

  
  
  
  

@gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
– Count Iblis
Sep 9 at 17:07

@gimusi Yes, that's a far quicker way to get to the answer (note that the sign of x/6 is positive, not negative).
– Count Iblis
Sep 9 at 17:07

ops yes of course!
– gimusi
Sep 9 at 17:22

ops yes of course!
– gimusi
Sep 9 at 17:22

I've added also a solution by that way!
– gimusi
2 days ago

I've added also a solution by that way!
– gimusi
2 days ago

add a comment | 

up vote
3
down vote

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$sin x = x-frac16 x^3+o(x^3) implies frac1sin x=frac 1xleft(1-frac16x^2+o(x^2)right)^-1=frac1x+frac16x+o(x)$$

therefore

$$left( frac1x^2 - frac1 sin^2 x right)
=left( frac1x + frac1 sin x right) left( frac1x - frac1 sin x right)=$$
$$=left(frac2x+frac16x+o(x)right) left( -frac16x+o(x)right)
=-frac13+o(1) to -frac13$$

share|cite|improve this answer

answered 2 days ago

gimusi

71.4k73786

add a comment | 

up vote
3
down vote

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$sin x = x-frac16 x^3+o(x^3) implies frac1sin x=frac 1xleft(1-frac16x^2+o(x^2)right)^-1=frac1x+frac16x+o(x)$$

therefore

$$left( frac1x^2 - frac1 sin^2 x right)
=left( frac1x + frac1 sin x right) left( frac1x - frac1 sin x right)=$$
$$=left(frac2x+frac16x+o(x)right) left( -frac16x+o(x)right)
=-frac13+o(1) to -frac13$$

share|cite|improve this answer

answered 2 days ago

gimusi

71.4k73786

add a comment | 

up vote
3
down vote

up vote
3
down vote

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$sin x = x-frac16 x^3+o(x^3) implies frac1sin x=frac 1xleft(1-frac16x^2+o(x^2)right)^-1=frac1x+frac16x+o(x)$$

therefore

$$left( frac1x^2 - frac1 sin^2 x right)
=left( frac1x + frac1 sin x right) left( frac1x - frac1 sin x right)=$$
$$=left(frac2x+frac16x+o(x)right) left( -frac16x+o(x)right)
=-frac13+o(1) to -frac13$$

share|cite|improve this answer

answered 2 days ago

gimusi

71.4k73786

As an alternative, following the idea by Count Iblis, we have that by Taylor expansion

$$sin x = x-frac16 x^3+o(x^3) implies frac1sin x=frac 1xleft(1-frac16x^2+o(x^2)right)^-1=frac1x+frac16x+o(x)$$

therefore

$$left( frac1x^2 - frac1 sin^2 x right)
=left( frac1x + frac1 sin x right) left( frac1x - frac1 sin x right)=$$
$$=left(frac2x+frac16x+o(x)right) left( -frac16x+o(x)right)
=-frac13+o(1) to -frac13$$

share|cite|improve this answer

answered 2 days ago

gimusi

71.4k73786

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

gimusi

71.4k73786

answered 2 days ago

gimusi

71.4k73786

answered 2 days ago

gimusi

71.4k73786

71.4k73786

add a comment | 

add a comment | 

up vote
2
down vote

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

---

What you can do instead (notice the asymmetry):

$$lim_xto0fracsin^2x-x^2x^2sin^2x=lim_xto0fracsin^2x-x^2x^4=lim_xto0fracsin x+xxlim_xto0fracsin x-xx^3
\=2lim_xto0fraccos x-13x^2=-2lim_xto0fracsin x6x=-frac13.$$

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edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

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Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

---

What you can do instead (notice the asymmetry):

$$lim_xto0fracsin^2x-x^2x^2sin^2x=lim_xto0fracsin^2x-x^2x^4=lim_xto0fracsin x+xxlim_xto0fracsin x-xx^3
\=2lim_xto0fraccos x-13x^2=-2lim_xto0fracsin x6x=-frac13.$$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

add a comment | 

up vote
2
down vote

up vote
2
down vote

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

---

What you can do instead (notice the asymmetry):

$$lim_xto0fracsin^2x-x^2x^2sin^2x=lim_xto0fracsin^2x-x^2x^4=lim_xto0fracsin x+xxlim_xto0fracsin x-xx^3
\=2lim_xto0fraccos x-13x^2=-2lim_xto0fracsin x6x=-frac13.$$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

Your mistake probably comes from your third row, because the left limit does not exist and you may not apply L'Hospital there (and the other limit is $0$).

---

What you can do instead (notice the asymmetry):

$$lim_xto0fracsin^2x-x^2x^2sin^2x=lim_xto0fracsin^2x-x^2x^4=lim_xto0fracsin x+xxlim_xto0fracsin x-xx^3
\=2lim_xto0fraccos x-13x^2=-2lim_xto0fracsin x6x=-frac13.$$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

share|cite|improve this answer

share|cite|improve this answer

edited 4 hours ago

edited 4 hours ago

edited 4 hours ago

answered 4 hours ago

Yves Daoust

114k665208

answered 4 hours ago

Yves Daoust

114k665208

answered 4 hours ago

Yves Daoust

114k665208

114k665208

add a comment | 

add a comment | 

up vote
1
down vote

$$lim_xto0frac(sinx+x)(sinx-x)xsinxcdot xsinx$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$lim_xto0fracsinx-xx^3=-frac16$$

$$lim_xto0fracx-tanxx^3=-frac13$$

$$lim_xto 0frace^x-1-xx^2=frac12$$

So using this,

$$lim_xto0fracx^2sin^2xcdot frac(sinx+x)xcdot frac(sinx-x)x^3$$

$$1cdot2cdot -frac16$$

$$-frac13$$

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answered 2 days ago

prog_SAHIL

1,464318

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up vote
1
down vote

$$lim_xto0frac(sinx+x)(sinx-x)xsinxcdot xsinx$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$lim_xto0fracsinx-xx^3=-frac16$$

$$lim_xto0fracx-tanxx^3=-frac13$$

$$lim_xto 0frace^x-1-xx^2=frac12$$

So using this,

$$lim_xto0fracx^2sin^2xcdot frac(sinx+x)xcdot frac(sinx-x)x^3$$

$$1cdot2cdot -frac16$$

$$-frac13$$

share|cite|improve this answer

answered 2 days ago

prog_SAHIL

1,464318

add a comment | 

up vote
1
down vote

up vote
1
down vote

$$lim_xto0frac(sinx+x)(sinx-x)xsinxcdot xsinx$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$lim_xto0fracsinx-xx^3=-frac16$$

$$lim_xto0fracx-tanxx^3=-frac13$$

$$lim_xto 0frace^x-1-xx^2=frac12$$

So using this,

$$lim_xto0fracx^2sin^2xcdot frac(sinx+x)xcdot frac(sinx-x)x^3$$

$$1cdot2cdot -frac16$$

$$-frac13$$

share|cite|improve this answer

answered 2 days ago

prog_SAHIL

1,464318

$$lim_xto0frac(sinx+x)(sinx-x)xsinxcdot xsinx$$

Here are some limits I remember that help me a lot, (easily derivable using L-Hopital)

$$lim_xto0fracsinx-xx^3=-frac16$$

$$lim_xto0fracx-tanxx^3=-frac13$$

$$lim_xto 0frace^x-1-xx^2=frac12$$

So using this,

$$lim_xto0fracx^2sin^2xcdot frac(sinx+x)xcdot frac(sinx-x)x^3$$

$$1cdot2cdot -frac16$$

$$-frac13$$

share|cite|improve this answer

answered 2 days ago

prog_SAHIL

1,464318

share|cite|improve this answer

share|cite|improve this answer

answered 2 days ago

prog_SAHIL

1,464318

answered 2 days ago

prog_SAHIL

1,464318

answered 2 days ago

prog_SAHIL

1,464318

1,464318

add a comment | 

add a comment | 

up vote
1
down vote

As noticed in the comments, we are allowed to proceed as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)= lim_x to 0 left( fracsin^2 x-x^2x^2sin^2 x right)=lim_x to 0 left( fracsin x+xxsin x right)left( fracsin x-xxsin x right)=ldots$$

but we are not allowed to proceed as follows

$$ldots=lim_x to 0 left( fracsin x+xxsin x right)lim_x to 0left( fracsin x-xxsin x right)=ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$ldots=lim_x to 0 frac cos x+1 sin x+xcos xcdot lim_x to 0 frac cos x-1 sin x+xcos x=ldots$$

and the LHS limit, in the form $frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ ldots=lim_x to 0 (cos x+1),lim_x to 0 frac cos x-1 (sin x+xcos x)^2=ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

---

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)
= lim_x to 0left(fracsin^2 x-x^2x^4cdotfracx^2sin^2 xright)
stackrel? = lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =ldots$$

and since, using l'Hopital for each part, we have

$$lim_x to 0fracsin^2 x-x^2x^4=lim_x to 0fracsin 2x-2x4x^3=lim_x to 0frac2cos 2x-212x^2=lim_x to 0frac-4sin 2x24x=lim_x to 0frac-8cos 2x24=-frac13$$

$$lim_x to 0fracx^2sin^2 x =lim_x to 0frac2xsin 2x =lim_x to 0frac22cos 2x =1$$

we see that the initial step is allowed and then we can conclude that

$$ldots= lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =-frac13cdot 1 =-frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $lim_x to 0fracsin x x=1$.

share|cite|improve this answer

edited 5 hours ago

answered yesterday

gimusi

71.4k73786

add a comment | 

up vote
1
down vote

As noticed in the comments, we are allowed to proceed as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)= lim_x to 0 left( fracsin^2 x-x^2x^2sin^2 x right)=lim_x to 0 left( fracsin x+xxsin x right)left( fracsin x-xxsin x right)=ldots$$

but we are not allowed to proceed as follows

$$ldots=lim_x to 0 left( fracsin x+xxsin x right)lim_x to 0left( fracsin x-xxsin x right)=ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$ldots=lim_x to 0 frac cos x+1 sin x+xcos xcdot lim_x to 0 frac cos x-1 sin x+xcos x=ldots$$

and the LHS limit, in the form $frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ ldots=lim_x to 0 (cos x+1),lim_x to 0 frac cos x-1 (sin x+xcos x)^2=ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

---

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)
= lim_x to 0left(fracsin^2 x-x^2x^4cdotfracx^2sin^2 xright)
stackrel? = lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =ldots$$

and since, using l'Hopital for each part, we have

$$lim_x to 0fracsin^2 x-x^2x^4=lim_x to 0fracsin 2x-2x4x^3=lim_x to 0frac2cos 2x-212x^2=lim_x to 0frac-4sin 2x24x=lim_x to 0frac-8cos 2x24=-frac13$$

$$lim_x to 0fracx^2sin^2 x =lim_x to 0frac2xsin 2x =lim_x to 0frac22cos 2x =1$$

we see that the initial step is allowed and then we can conclude that

$$ldots= lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =-frac13cdot 1 =-frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $lim_x to 0fracsin x x=1$.

share|cite|improve this answer

edited 5 hours ago

answered yesterday

gimusi

71.4k73786

add a comment | 

up vote
1
down vote

up vote
1
down vote

As noticed in the comments, we are allowed to proceed as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)= lim_x to 0 left( fracsin^2 x-x^2x^2sin^2 x right)=lim_x to 0 left( fracsin x+xxsin x right)left( fracsin x-xxsin x right)=ldots$$

but we are not allowed to proceed as follows

$$ldots=lim_x to 0 left( fracsin x+xxsin x right)lim_x to 0left( fracsin x-xxsin x right)=ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$ldots=lim_x to 0 frac cos x+1 sin x+xcos xcdot lim_x to 0 frac cos x-1 sin x+xcos x=ldots$$

and the LHS limit, in the form $frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ ldots=lim_x to 0 (cos x+1),lim_x to 0 frac cos x-1 (sin x+xcos x)^2=ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

---

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)
= lim_x to 0left(fracsin^2 x-x^2x^4cdotfracx^2sin^2 xright)
stackrel? = lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =ldots$$

and since, using l'Hopital for each part, we have

$$lim_x to 0fracsin^2 x-x^2x^4=lim_x to 0fracsin 2x-2x4x^3=lim_x to 0frac2cos 2x-212x^2=lim_x to 0frac-4sin 2x24x=lim_x to 0frac-8cos 2x24=-frac13$$

$$lim_x to 0fracx^2sin^2 x =lim_x to 0frac2xsin 2x =lim_x to 0frac22cos 2x =1$$

we see that the initial step is allowed and then we can conclude that

$$ldots= lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =-frac13cdot 1 =-frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $lim_x to 0fracsin x x=1$.

share|cite|improve this answer

edited 5 hours ago

answered yesterday

gimusi

71.4k73786

As noticed in the comments, we are allowed to proceed as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)= lim_x to 0 left( fracsin^2 x-x^2x^2sin^2 x right)=lim_x to 0 left( fracsin x+xxsin x right)left( fracsin x-xxsin x right)=ldots$$

but we are not allowed to proceed as follows

$$ldots=lim_x to 0 left( fracsin x+xxsin x right)lim_x to 0left( fracsin x-xxsin x right)=ldots$$

when one or both limits do not exist or the product leads to an undefined expression.

Notably in that case by l'Hopital we obtain

$$ldots=lim_x to 0 frac cos x+1 sin x+xcos xcdot lim_x to 0 frac cos x-1 sin x+xcos x=ldots$$

and the LHS limit, in the form $frac 2 0$, doesn't exist while the RHS limit is equal to zero.

Therefore the initial step in that case doesn't work.

Note that in any case also the following step

$$ ldots=lim_x to 0 (cos x+1),lim_x to 0 frac cos x-1 (sin x+xcos x)^2=ldots$$

is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.

See also the related Analyzing limits problem Calculus (tell me where I'm wrong).

---

In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows

$$lim_x to 0 left( frac1x^2 - frac1 sin^2 x right)
= lim_x to 0left(fracsin^2 x-x^2x^4cdotfracx^2sin^2 xright)
stackrel? = lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =ldots$$

and since, using l'Hopital for each part, we have

$$lim_x to 0fracsin^2 x-x^2x^4=lim_x to 0fracsin 2x-2x4x^3=lim_x to 0frac2cos 2x-212x^2=lim_x to 0frac-4sin 2x24x=lim_x to 0frac-8cos 2x24=-frac13$$

$$lim_x to 0fracx^2sin^2 x =lim_x to 0frac2xsin 2x =lim_x to 0frac22cos 2x =1$$

we see that the initial step is allowed and then we can conclude that

$$ldots= lim_x to 0fracsin^2 x-x^2x^4cdotlim_x to 0fracx^2sin^2 x =-frac13cdot 1 =-frac13$$

Note finally that some intermediate step can be highly simplified using the standard limit $lim_x to 0fracsin x x=1$.

share|cite|improve this answer

edited 5 hours ago

answered yesterday

gimusi

71.4k73786

share|cite|improve this answer

share|cite|improve this answer

edited 5 hours ago

edited 5 hours ago

edited 5 hours ago

answered yesterday

gimusi

71.4k73786

answered yesterday

gimusi

71.4k73786

answered yesterday

gimusi

71.4k73786

71.4k73786

add a comment | 

add a comment | 

 

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