Mixing
Doubt on limits evaluation.
Clash Royale CLAN TAG#URR8PPP
up vote
8
down vote
favorite
Can anyone tell me what am I doing wrong here.
The limit provided is$$lim_x to 0dfracxe^x-ln(1+x)x²$$
Method 1 (using standard limits)
$$= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
$$= dfracdisplaystyle lim_x to 0e^x-lim_x to 0 1displaystyle lim_x to 0x$$
$$= lim_x to 0 dfrace^x-1x$$
$$= 1$$
Method 2 (using Maclaurin series )
$$lim_x to 0dfracxe^x-ln(1+x)x²$$
$$= dfrac32$$
Even with L'Hopital rule I get $dfrac32$. Then what's wrong with method 1.
Some limit's properties
limits
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
8
down vote
favorite
Can anyone tell me what am I doing wrong here.
The limit provided is$$lim_x to 0dfracxe^x-ln(1+x)x²$$
Method 1 (using standard limits)
$$= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
$$= dfracdisplaystyle lim_x to 0e^x-lim_x to 0 1displaystyle lim_x to 0x$$
$$= lim_x to 0 dfrace^x-1x$$
$$= 1$$
Method 2 (using Maclaurin series )
$$lim_x to 0dfracxe^x-ln(1+x)x²$$
$$= dfrac32$$
Even with L'Hopital rule I get $dfrac32$. Then what's wrong with method 1.
Some limit's properties
limits
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
1
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
1
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Can anyone tell me what am I doing wrong here.
The limit provided is$$lim_x to 0dfracxe^x-ln(1+x)x²$$
Method 1 (using standard limits)
$$= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
$$= dfracdisplaystyle lim_x to 0e^x-lim_x to 0 1displaystyle lim_x to 0x$$
$$= lim_x to 0 dfrace^x-1x$$
$$= 1$$
Method 2 (using Maclaurin series )
$$lim_x to 0dfracxe^x-ln(1+x)x²$$
$$= dfrac32$$
Even with L'Hopital rule I get $dfrac32$. Then what's wrong with method 1.
Some limit's properties
limits
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Can anyone tell me what am I doing wrong here.
The limit provided is$$lim_x to 0dfracxe^x-ln(1+x)x²$$
Method 1 (using standard limits)
$$= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
$$= dfracdisplaystyle lim_x to 0e^x-lim_x to 0 1displaystyle lim_x to 0x$$
$$= lim_x to 0 dfrace^x-1x$$
$$= 1$$
Method 2 (using Maclaurin series )
$$lim_x to 0dfracxe^x-ln(1+x)x²$$
$$= dfrac32$$
Even with L'Hopital rule I get $dfrac32$. Then what's wrong with method 1.
Some limit's properties
limits
limits
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 23 hours ago
asked yesterday
Loop Back
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Loop Back
437
asked yesterday
Loop Back
437
437
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Loop Back is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
1
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
1
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago
add a comment |Â
5
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
1
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
1
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago
5
5
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
1
1
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
1
1
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
down vote
accepted
The following step
$$lim_x to 0dfracxe^x-ln(1+x)x²= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
is not allowed since it leads to an undefined expression $frac 0 0$.
See also the related
- Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $
- Analyzing limits problem Calculus (tell me where I'm wrong).
answered yesterday
gimusi
71.6k73786
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
add a comment |Â
up vote
10
down vote
The error is quite simply described.
Theorem. If $lim_xto af(x)=l$ and $lim_xto ag(x)=m$ and $mne0$, then
$$
lim_xto afracf(x)g(x)=fraclm
$$
This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $lim_xto0x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.
answered 23 hours ago
egreg
166k1180189
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
add a comment |Â
up vote
4
down vote
ORIGINAL ANSWER
You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.
UPDATE.
I kind of misread the question. The real issue is that the $
requireenclose
enclosehorizontalstriketextlimit is an indeterminate form
$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write
$$
lim_xto 0 frac 1-cos(x) x^2 = frac lim_xto 01 - lim_x to 0 cos(x) lim_xto 0 x^2 = frac 00,
$$
which is absurd.
Appendices
Some discussions with @LoopBack & @HenryLee are posted here for convenience.
I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$lim_x to infty x²+x =+infty,$$ here $ ell$ and $m$ both are $infty$.
$lim_xto +infty x^2 - x^2 = 0$ but $lim_x to +infty x^2 -x = +infty$, also $lim_n n - (-1)^n$ [Let $g(x) = (-1)^lfloor xrfloor lfloor x rfloor, f(x) = lfloor x rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+infty + (+infty)$ could be accepted, but the problem is $(+infty) - (+infty)$. Also I do not think textbooks allows $infty$ in the theorem involving arithmetic operations.
Can you give me an example where $$displaystyle lim_x to cf(x) + g(x) =displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$$ and $displaystyle lim_x to cf(x) + g(x) $ is indeterminant whereas $displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3
If $lim f, lim g$ is not indeterminate [not including $infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $lim(fpm g)$ is also not indeterminate. (ii)(iii) are no way useless here.
UPDATE 2
My terminology was wrong. The thing matters is that $lim f, lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].
answered yesterday
xbh
3,345320
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
 |Â
show 6 more comments
up vote
-1
down vote
$$lim_x to 0fracxe^x-ln(1+x)x^2=lim_x to 0fracxe^xx^2-lim_x to 0fracln(1+x)x^2=lim_x to 0frace^xx-lim_x to 0fracfrac11+x2x=lim_x to 0e^x-lim_x to 0frac12x+2x^2=1-lim_x to 0frac12x+2x^2$$
this final limit appears to approach infinity so the answer is $+infty$ or $-infty$ depending on whether we use $0^+$ or $0^-$
answered 23 hours ago
Henry Lee
62310
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The following step
$$lim_x to 0dfracxe^x-ln(1+x)x²= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
is not allowed since it leads to an undefined expression $frac 0 0$.
See also the related
- Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $
- Analyzing limits problem Calculus (tell me where I'm wrong).
answered yesterday
gimusi
71.6k73786
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
add a comment |Â
up vote
7
down vote
accepted
The following step
$$lim_x to 0dfracxe^x-ln(1+x)x²= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
is not allowed since it leads to an undefined expression $frac 0 0$.
See also the related
- Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $
- Analyzing limits problem Calculus (tell me where I'm wrong).
answered yesterday
gimusi
71.6k73786
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The following step
$$lim_x to 0dfracxe^x-ln(1+x)x²= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
is not allowed since it leads to an undefined expression $frac 0 0$.
See also the related
- Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $
- Analyzing limits problem Calculus (tell me where I'm wrong).
answered yesterday
gimusi
71.6k73786
The following step
$$lim_x to 0dfracxe^x-ln(1+x)x²= dfracdisplaystyle lim_x to 0dfracxe^xx-lim_x to 0dfracln(1+x)xdisplaystyle lim_x to 0x$$
is not allowed since it leads to an undefined expression $frac 0 0$.
See also the related
- Evaluate $ lim_x to 0 left( frac1x^2 - frac1 sin^2 x right) $
- Analyzing limits problem Calculus (tell me where I'm wrong).
answered yesterday
gimusi
71.6k73786
edited 22 hours ago
answered yesterday
gimusi
71.6k73786
answered yesterday
gimusi
71.6k73786
answered yesterday
gimusi
71.6k73786
71.6k73786
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
add a comment |Â
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
6
6
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
It leads to an undefined expression, rather than an “indeterminate formâ€Â: it is always disallowed to divide by $0$.
– egreg
23 hours ago
1
1
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
@egreg Thanks for your clarification. My answer is somehow not correct now.
– xbh
23 hours ago
1
1
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
@egreg Yes you are absolutely right, I fix that! Thanks
– gimusi
22 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
+1 for the second link.
– Paramanand Singh
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
@ParamanandSingh It's a main reference for me! ;)
– gimusi
4 hours ago
add a comment |Â
up vote
10
down vote
The error is quite simply described.
Theorem. If $lim_xto af(x)=l$ and $lim_xto ag(x)=m$ and $mne0$, then
$$
lim_xto afracf(x)g(x)=fraclm
$$
This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $lim_xto0x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.
answered 23 hours ago
egreg
166k1180189
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
add a comment |Â
up vote
10
down vote
The error is quite simply described.
Theorem. If $lim_xto af(x)=l$ and $lim_xto ag(x)=m$ and $mne0$, then
$$
lim_xto afracf(x)g(x)=fraclm
$$
This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $lim_xto0x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.
answered 23 hours ago
egreg
166k1180189
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
add a comment |Â
up vote
10
down vote
up vote
10
down vote
The error is quite simply described.
Theorem. If $lim_xto af(x)=l$ and $lim_xto ag(x)=m$ and $mne0$, then
$$
lim_xto afracf(x)g(x)=fraclm
$$
This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $lim_xto0x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.
answered 23 hours ago
egreg
166k1180189
The error is quite simply described.
Theorem. If $lim_xto af(x)=l$ and $lim_xto ag(x)=m$ and $mne0$, then
$$
lim_xto afracf(x)g(x)=fraclm
$$
This is a correct statement. But you cannot apply it when not all the hypotheses are satisfied.
In your case, $lim_xto0x=0$, so the theorem cannot be applied.
Why can't it? Because division by $0$ is not allowed.
answered 23 hours ago
egreg
166k1180189
answered 23 hours ago
egreg
166k1180189
answered 23 hours ago
egreg
166k1180189
answered 23 hours ago
egreg
166k1180189
166k1180189
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
add a comment |Â
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
1
1
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
Thank you next time I'll be careful enough
– Loop Back
23 hours ago
add a comment |Â
up vote
4
down vote
ORIGINAL ANSWER
You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.
UPDATE.
I kind of misread the question. The real issue is that the $
requireenclose
enclosehorizontalstriketextlimit is an indeterminate form
$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write
$$
lim_xto 0 frac 1-cos(x) x^2 = frac lim_xto 01 - lim_x to 0 cos(x) lim_xto 0 x^2 = frac 00,
$$
which is absurd.
Appendices
Some discussions with @LoopBack & @HenryLee are posted here for convenience.
I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$lim_x to infty x²+x =+infty,$$ here $ ell$ and $m$ both are $infty$.
$lim_xto +infty x^2 - x^2 = 0$ but $lim_x to +infty x^2 -x = +infty$, also $lim_n n - (-1)^n$ [Let $g(x) = (-1)^lfloor xrfloor lfloor x rfloor, f(x) = lfloor x rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+infty + (+infty)$ could be accepted, but the problem is $(+infty) - (+infty)$. Also I do not think textbooks allows $infty$ in the theorem involving arithmetic operations.
Can you give me an example where $$displaystyle lim_x to cf(x) + g(x) =displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$$ and $displaystyle lim_x to cf(x) + g(x) $ is indeterminant whereas $displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3
If $lim f, lim g$ is not indeterminate [not including $infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $lim(fpm g)$ is also not indeterminate. (ii)(iii) are no way useless here.
UPDATE 2
My terminology was wrong. The thing matters is that $lim f, lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].
answered yesterday
xbh
3,345320
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
 |Â
show 6 more comments
up vote
4
down vote
ORIGINAL ANSWER
You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.
UPDATE.
I kind of misread the question. The real issue is that the $
requireenclose
enclosehorizontalstriketextlimit is an indeterminate form
$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write
$$
lim_xto 0 frac 1-cos(x) x^2 = frac lim_xto 01 - lim_x to 0 cos(x) lim_xto 0 x^2 = frac 00,
$$
which is absurd.
Appendices
Some discussions with @LoopBack & @HenryLee are posted here for convenience.
I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$lim_x to infty x²+x =+infty,$$ here $ ell$ and $m$ both are $infty$.
$lim_xto +infty x^2 - x^2 = 0$ but $lim_x to +infty x^2 -x = +infty$, also $lim_n n - (-1)^n$ [Let $g(x) = (-1)^lfloor xrfloor lfloor x rfloor, f(x) = lfloor x rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+infty + (+infty)$ could be accepted, but the problem is $(+infty) - (+infty)$. Also I do not think textbooks allows $infty$ in the theorem involving arithmetic operations.
Can you give me an example where $$displaystyle lim_x to cf(x) + g(x) =displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$$ and $displaystyle lim_x to cf(x) + g(x) $ is indeterminant whereas $displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3
If $lim f, lim g$ is not indeterminate [not including $infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $lim(fpm g)$ is also not indeterminate. (ii)(iii) are no way useless here.
UPDATE 2
My terminology was wrong. The thing matters is that $lim f, lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].
answered yesterday
xbh
3,345320
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
 |Â
show 6 more comments
up vote
4
down vote
up vote
4
down vote
ORIGINAL ANSWER
You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.
UPDATE.
I kind of misread the question. The real issue is that the $
requireenclose
enclosehorizontalstriketextlimit is an indeterminate form
$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write
$$
lim_xto 0 frac 1-cos(x) x^2 = frac lim_xto 01 - lim_x to 0 cos(x) lim_xto 0 x^2 = frac 00,
$$
which is absurd.
Appendices
Some discussions with @LoopBack & @HenryLee are posted here for convenience.
I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$lim_x to infty x²+x =+infty,$$ here $ ell$ and $m$ both are $infty$.
$lim_xto +infty x^2 - x^2 = 0$ but $lim_x to +infty x^2 -x = +infty$, also $lim_n n - (-1)^n$ [Let $g(x) = (-1)^lfloor xrfloor lfloor x rfloor, f(x) = lfloor x rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+infty + (+infty)$ could be accepted, but the problem is $(+infty) - (+infty)$. Also I do not think textbooks allows $infty$ in the theorem involving arithmetic operations.
Can you give me an example where $$displaystyle lim_x to cf(x) + g(x) =displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$$ and $displaystyle lim_x to cf(x) + g(x) $ is indeterminant whereas $displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3
If $lim f, lim g$ is not indeterminate [not including $infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $lim(fpm g)$ is also not indeterminate. (ii)(iii) are no way useless here.
UPDATE 2
My terminology was wrong. The thing matters is that $lim f, lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].
answered yesterday
xbh
3,345320
ORIGINAL ANSWER
You can only do that in method 1 when each limit exists. Since the law of arithmetic operation holds only when the limit of each term exists. Counterexample could be your method 1.
UPDATE.
I kind of misread the question. The real issue is that the $
requireenclose
enclosehorizontalstriketextlimit is an indeterminate form
$ the denominator tends to zero [thanks to @egreg]. Similar to what OP has used, i might also write
$$
lim_xto 0 frac 1-cos(x) x^2 = frac lim_xto 01 - lim_x to 0 cos(x) lim_xto 0 x^2 = frac 00,
$$
which is absurd.
Appendices
Some discussions with @LoopBack & @HenryLee are posted here for convenience.
I do not see why the expression cannot be split into multiple parts, and if it cannot are there cases when it is possible? /// Who said that $l,m$ should be real? Take an example $$lim_x to infty x²+x =+infty,$$ here $ ell$ and $m$ both are $infty$.
$lim_xto +infty x^2 - x^2 = 0$ but $lim_x to +infty x^2 -x = +infty$, also $lim_n n - (-1)^n$ [Let $g(x) = (-1)^lfloor xrfloor lfloor x rfloor, f(x) = lfloor x rfloor$] simply does not exist. Therefore we generally cannot split the expression into two parts without any justification. The type $+infty + (+infty)$ could be accepted, but the problem is $(+infty) - (+infty)$. Also I do not think textbooks allows $infty$ in the theorem involving arithmetic operations.
Can you give me an example where $$displaystyle lim_x to cf(x) + g(x) =displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$$ and $displaystyle lim_x to cf(x) + g(x) $ is indeterminant whereas $displaystyle lim_x to cf(x) +displaystyle lim_x to c g(x)$ is not. I don't think so there is such expression. So what is the use of property no. 2 and 3
If $lim f, lim g$ is not indeterminate [not including $infty$], then you could use (ii)(iii). By virtue of (ii)(iii) $lim(fpm g)$ is also not indeterminate. (ii)(iii) are no way useless here.
UPDATE 2
My terminology was wrong. The thing matters is that $lim f, lim g$ exists, not concerning whether $f,g$ is in indeterminate form. If we split into two parts and both of them are, say $0/0$-form, but both of them exists [again, no $infty$] after investigation, then clearly the splitting operation is justified. My point is, logically speaking, that we should verify the existence first, then break the original expression into several parts to handle according to the (ii) & (iii) [although this never means to do the process chronologically, i.e. you could verify after completing the computation].
answered yesterday
xbh
3,345320
edited 20 hours ago
answered yesterday
xbh
3,345320
answered yesterday
xbh
3,345320
answered yesterday
xbh
3,345320
3,345320
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
 |Â
show 6 more comments
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
1
1
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
What do mean by "You can only do that in method 1 when each limit exists.". What thing plz elaborate.
– Loop Back
yesterday
1
1
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
Sorry, that's not the issue. But could you justify each step in method 1 in your post?
– xbh
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What do you want me to elaborate I have used standard limits for $dfracln(1+x)x$ and for $lim_x to 0 dfrace^x-1x$
– Loop Back
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
What theorem guarantees that you could push $lim$ symbol inside?
– xbh
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
Aren't they the basic properties of limit. Let me upload a pic.
– Loop Back
yesterday
 |Â
show 6 more comments
up vote
-1
down vote
$$lim_x to 0fracxe^x-ln(1+x)x^2=lim_x to 0fracxe^xx^2-lim_x to 0fracln(1+x)x^2=lim_x to 0frace^xx-lim_x to 0fracfrac11+x2x=lim_x to 0e^x-lim_x to 0frac12x+2x^2=1-lim_x to 0frac12x+2x^2$$
this final limit appears to approach infinity so the answer is $+infty$ or $-infty$ depending on whether we use $0^+$ or $0^-$
answered 23 hours ago
Henry Lee
62310
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
add a comment |Â
up vote
-1
down vote
$$lim_x to 0fracxe^x-ln(1+x)x^2=lim_x to 0fracxe^xx^2-lim_x to 0fracln(1+x)x^2=lim_x to 0frace^xx-lim_x to 0fracfrac11+x2x=lim_x to 0e^x-lim_x to 0frac12x+2x^2=1-lim_x to 0frac12x+2x^2$$
this final limit appears to approach infinity so the answer is $+infty$ or $-infty$ depending on whether we use $0^+$ or $0^-$
answered 23 hours ago
Henry Lee
62310
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
$$lim_x to 0fracxe^x-ln(1+x)x^2=lim_x to 0fracxe^xx^2-lim_x to 0fracln(1+x)x^2=lim_x to 0frace^xx-lim_x to 0fracfrac11+x2x=lim_x to 0e^x-lim_x to 0frac12x+2x^2=1-lim_x to 0frac12x+2x^2$$
this final limit appears to approach infinity so the answer is $+infty$ or $-infty$ depending on whether we use $0^+$ or $0^-$
answered 23 hours ago
Henry Lee
62310
$$lim_x to 0fracxe^x-ln(1+x)x^2=lim_x to 0fracxe^xx^2-lim_x to 0fracln(1+x)x^2=lim_x to 0frace^xx-lim_x to 0fracfrac11+x2x=lim_x to 0e^x-lim_x to 0frac12x+2x^2=1-lim_x to 0frac12x+2x^2$$
this final limit appears to approach infinity so the answer is $+infty$ or $-infty$ depending on whether we use $0^+$ or $0^-$
answered 23 hours ago
Henry Lee
62310
answered 23 hours ago
Henry Lee
62310
answered 23 hours ago
Henry Lee
62310
answered 23 hours ago
Henry Lee
62310
62310
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
add a comment |Â
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
21 hours ago
add a comment |Â
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5
You put in and out $lim $ inside expressions without knowing that these limits exist.
– dmtri
yesterday
1
@dmtri I'll be conscious next time. Failure are stepping stones to success
– Loop Back
23 hours ago
1
In Method 1 you turned the denominator from $x^2$ to $x$. Was that intentional? It might just be a transcription issue.
– Brian J
21 hours ago
Actually I took one of the $x$ from $x²$ to the numerator, maybe you need to go through it once again
– Loop Back
21 hours ago
Note that, almost always, you can just use a Taylor series expansion to get what you'd get with de l'hopital but without easily falling trap from this kind of mistake. In any case, de L'Hopital theorem requires your derivative of the deniominator to never be $0$, and in fact here you have that the derivative of $x^2$ is $2x$ which is $0$ for $x rightarrow 0$.
– Bakuriu
15 hours ago