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Any idea how to solve this limit?
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$$lim_xto 0 fracsqrt1-cos(x^2)1-cos(x)$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $sqrt2$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $cos$.
Then i tried L'Hopital because it is $frac00$ and still that $cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
limits
edited 2 hours ago
Jakobian
2,009518
asked 2 hours ago
jkuzmanovik
111
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jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
$$lim_xto 0 fracsqrt1-cos(x^2)1-cos(x)$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $sqrt2$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $cos$.
Then i tried L'Hopital because it is $frac00$ and still that $cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
limits
edited 2 hours ago
Jakobian
2,009518
asked 2 hours ago
jkuzmanovik
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$lim_xto 0 fracsqrt1-cos(x^2)1-cos(x)$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $sqrt2$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $cos$.
Then i tried L'Hopital because it is $frac00$ and still that $cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
limits
edited 2 hours ago
Jakobian
2,009518
asked 2 hours ago
jkuzmanovik
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$lim_xto 0 fracsqrt1-cos(x^2)1-cos(x)$$
I am trying to solve this limit for 2 days, but still cant find the solution which is $sqrt2$ (that's what is written in the solution sheet)
I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $cos$.
Then i tried L'Hopital because it is $frac00$ and still that $cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.
So can someone help me how to solve this?
Thank you.
limits
limits
edited 2 hours ago
Jakobian
2,009518
asked 2 hours ago
jkuzmanovik
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Jakobian
2,009518
asked 2 hours ago
jkuzmanovik
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Jakobian
2,009518
edited 2 hours ago
Jakobian
2,009518
edited 2 hours ago
Jakobian
2,009518
2,009518
asked 2 hours ago
jkuzmanovik
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
jkuzmanovik
111
asked 2 hours ago
jkuzmanovik
111
111
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
jkuzmanovik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago
add a comment |Â
You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago
You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
An approach based on the fact that
$$
lim_uto0 frac1-cos uu^2 = frac12 tag1
$$
which is a standard limit (and equivalent to a Taylor expansion of $cos$ to order $2$ at $0$).
$$
fracsqrt1-cos(x^2)1-cos x
= sqrtfrac1-cos(x^2)x^4cdotfracx^21-cos x xrightarrow[xto0]sqrtfrac12cdotfrac1frac12 = sqrt2,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2to 0$ when $xto 0$).
To be completely precise: we used (1) twice, and the continuity of both $sqrtcdot$ and the inverse function at $1/2$ to have $$lim_xto 0 sqrtg(x)frac1h(x) = lim_xto 0sqrtg(x)lim_xto 0frac1h(x) = sqrtlim_xto 0g(x)frac1lim_xto 0h(x)$$
answered 2 hours ago
Clement C.
47.7k33783
add a comment |Â
up vote
1
down vote
By L'Hopital's rule, we have
$$beginaligned
lim_x to 0frac1 - cos(x^2)(1 - cos(x))^2
&= lim_x to 0frac2xsin(x^2)2(1 - cos(x))sin(x) \
&= lim_x to 0fracxsin(x)fracsin(x^2)1 - cos(x) \
&= lim_x to 0fracsin(x^2)1 - cos(x) \
endaligned$$
where we have used that
$$lim_x to 0fracxsin(x) = 1$$
Now we can apply L'Hopital's rule again to obtain
$$beginaligned
lim_x to 0fracsin(x^2)1 - cos(x)
&= lim_x to 0 frac2xcos(x^2)sin(x) \
&= lim_x to 0 fracxsin(x) 2cos(x^2) \
&= lim_x to 0 2cos(x^2) \
&= 2
endaligned$$
Summarizing what we have so far,
$$lim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 = 2$$
Now take the square root of both sides and use the fact that $sqrt(cdot)$ is continuous at zero (note that we approach only from the right) to conclude that
$$
beginaligned
lim_x to 0fracsqrt1 - cos(x^2)1 - cos(x)
&= lim_x to 0sqrtfrac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrtlim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrt2
endaligned$$
as desired.
answered 2 hours ago
Bungo
13.2k22044
add a comment |Â
up vote
1
down vote
if you multiply and divide by both conjugates you get:
$$lim_x to 0 frac(1+cosx)sin(x^2)sqrt1+cos(x^2) space sin^2x$$
Amplifying to get limit of the form $fracsinxx :$
$$=lim_x to 0 frac(1+cosx)sqrt1+cosx^2 space fracsin(x^2)x^2 space fracx^2sin^2x=frac2sqrt2*1*1^2=sqrt2$$
answered 2 hours ago
Villa
44310
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
add a comment |Â
up vote
1
down vote
beginalign*
lim_xto 0fracsqrt1-cos(x^2)1-cos(x)
&=lim_xto 0fracsqrt1-cos(x^2)cdot(1+cos x)1-cos^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)sin^2(x)\[10pt]
&=2lim_xto 0fracsqrt1-cos(x^2)x^2cdotfracx^2sin^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)x^2
stackrely:=x^2=2lim_yto 0^+fracsqrt1-cos(y)y\[10pt]
&=2lim_yto 0^+sqrtfrac1-cos(y)2cdotfracsqrt 2y
=2lim_yto 0^2sinleft(frac y2right)cdotfracsqrt 2y
=sqrt 2lim_yto 0^+sinleft(frac y2right)cdotfrac2y=boldsymbolsqrt 2
endalign*
answered 2 hours ago
Pavel R.
25126
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
An approach based on the fact that
$$
lim_uto0 frac1-cos uu^2 = frac12 tag1
$$
which is a standard limit (and equivalent to a Taylor expansion of $cos$ to order $2$ at $0$).
$$
fracsqrt1-cos(x^2)1-cos x
= sqrtfrac1-cos(x^2)x^4cdotfracx^21-cos x xrightarrow[xto0]sqrtfrac12cdotfrac1frac12 = sqrt2,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2to 0$ when $xto 0$).
To be completely precise: we used (1) twice, and the continuity of both $sqrtcdot$ and the inverse function at $1/2$ to have $$lim_xto 0 sqrtg(x)frac1h(x) = lim_xto 0sqrtg(x)lim_xto 0frac1h(x) = sqrtlim_xto 0g(x)frac1lim_xto 0h(x)$$
answered 2 hours ago
Clement C.
47.7k33783
add a comment |Â
up vote
1
down vote
An approach based on the fact that
$$
lim_uto0 frac1-cos uu^2 = frac12 tag1
$$
which is a standard limit (and equivalent to a Taylor expansion of $cos$ to order $2$ at $0$).
$$
fracsqrt1-cos(x^2)1-cos x
= sqrtfrac1-cos(x^2)x^4cdotfracx^21-cos x xrightarrow[xto0]sqrtfrac12cdotfrac1frac12 = sqrt2,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2to 0$ when $xto 0$).
To be completely precise: we used (1) twice, and the continuity of both $sqrtcdot$ and the inverse function at $1/2$ to have $$lim_xto 0 sqrtg(x)frac1h(x) = lim_xto 0sqrtg(x)lim_xto 0frac1h(x) = sqrtlim_xto 0g(x)frac1lim_xto 0h(x)$$
answered 2 hours ago
Clement C.
47.7k33783
add a comment |Â
up vote
1
down vote
up vote
1
down vote
An approach based on the fact that
$$
lim_uto0 frac1-cos uu^2 = frac12 tag1
$$
which is a standard limit (and equivalent to a Taylor expansion of $cos$ to order $2$ at $0$).
$$
fracsqrt1-cos(x^2)1-cos x
= sqrtfrac1-cos(x^2)x^4cdotfracx^21-cos x xrightarrow[xto0]sqrtfrac12cdotfrac1frac12 = sqrt2,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2to 0$ when $xto 0$).
To be completely precise: we used (1) twice, and the continuity of both $sqrtcdot$ and the inverse function at $1/2$ to have $$lim_xto 0 sqrtg(x)frac1h(x) = lim_xto 0sqrtg(x)lim_xto 0frac1h(x) = sqrtlim_xto 0g(x)frac1lim_xto 0h(x)$$
answered 2 hours ago
Clement C.
47.7k33783
An approach based on the fact that
$$
lim_uto0 frac1-cos uu^2 = frac12 tag1
$$
which is a standard limit (and equivalent to a Taylor expansion of $cos$ to order $2$ at $0$).
$$
fracsqrt1-cos(x^2)1-cos x
= sqrtfrac1-cos(x^2)x^4cdotfracx^21-cos x xrightarrow[xto0]sqrtfrac12cdotfrac1frac12 = sqrt2,,
$$
applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2to 0$ when $xto 0$).
To be completely precise: we used (1) twice, and the continuity of both $sqrtcdot$ and the inverse function at $1/2$ to have $$lim_xto 0 sqrtg(x)frac1h(x) = lim_xto 0sqrtg(x)lim_xto 0frac1h(x) = sqrtlim_xto 0g(x)frac1lim_xto 0h(x)$$
answered 2 hours ago
Clement C.
47.7k33783
answered 2 hours ago
Clement C.
47.7k33783
answered 2 hours ago
Clement C.
47.7k33783
answered 2 hours ago
Clement C.
47.7k33783
47.7k33783
add a comment |Â
add a comment |Â
up vote
1
down vote
By L'Hopital's rule, we have
$$beginaligned
lim_x to 0frac1 - cos(x^2)(1 - cos(x))^2
&= lim_x to 0frac2xsin(x^2)2(1 - cos(x))sin(x) \
&= lim_x to 0fracxsin(x)fracsin(x^2)1 - cos(x) \
&= lim_x to 0fracsin(x^2)1 - cos(x) \
endaligned$$
where we have used that
$$lim_x to 0fracxsin(x) = 1$$
Now we can apply L'Hopital's rule again to obtain
$$beginaligned
lim_x to 0fracsin(x^2)1 - cos(x)
&= lim_x to 0 frac2xcos(x^2)sin(x) \
&= lim_x to 0 fracxsin(x) 2cos(x^2) \
&= lim_x to 0 2cos(x^2) \
&= 2
endaligned$$
Summarizing what we have so far,
$$lim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 = 2$$
Now take the square root of both sides and use the fact that $sqrt(cdot)$ is continuous at zero (note that we approach only from the right) to conclude that
$$
beginaligned
lim_x to 0fracsqrt1 - cos(x^2)1 - cos(x)
&= lim_x to 0sqrtfrac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrtlim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrt2
endaligned$$
as desired.
answered 2 hours ago
Bungo
13.2k22044
add a comment |Â
up vote
1
down vote
By L'Hopital's rule, we have
$$beginaligned
lim_x to 0frac1 - cos(x^2)(1 - cos(x))^2
&= lim_x to 0frac2xsin(x^2)2(1 - cos(x))sin(x) \
&= lim_x to 0fracxsin(x)fracsin(x^2)1 - cos(x) \
&= lim_x to 0fracsin(x^2)1 - cos(x) \
endaligned$$
where we have used that
$$lim_x to 0fracxsin(x) = 1$$
Now we can apply L'Hopital's rule again to obtain
$$beginaligned
lim_x to 0fracsin(x^2)1 - cos(x)
&= lim_x to 0 frac2xcos(x^2)sin(x) \
&= lim_x to 0 fracxsin(x) 2cos(x^2) \
&= lim_x to 0 2cos(x^2) \
&= 2
endaligned$$
Summarizing what we have so far,
$$lim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 = 2$$
Now take the square root of both sides and use the fact that $sqrt(cdot)$ is continuous at zero (note that we approach only from the right) to conclude that
$$
beginaligned
lim_x to 0fracsqrt1 - cos(x^2)1 - cos(x)
&= lim_x to 0sqrtfrac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrtlim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrt2
endaligned$$
as desired.
answered 2 hours ago
Bungo
13.2k22044
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By L'Hopital's rule, we have
$$beginaligned
lim_x to 0frac1 - cos(x^2)(1 - cos(x))^2
&= lim_x to 0frac2xsin(x^2)2(1 - cos(x))sin(x) \
&= lim_x to 0fracxsin(x)fracsin(x^2)1 - cos(x) \
&= lim_x to 0fracsin(x^2)1 - cos(x) \
endaligned$$
where we have used that
$$lim_x to 0fracxsin(x) = 1$$
Now we can apply L'Hopital's rule again to obtain
$$beginaligned
lim_x to 0fracsin(x^2)1 - cos(x)
&= lim_x to 0 frac2xcos(x^2)sin(x) \
&= lim_x to 0 fracxsin(x) 2cos(x^2) \
&= lim_x to 0 2cos(x^2) \
&= 2
endaligned$$
Summarizing what we have so far,
$$lim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 = 2$$
Now take the square root of both sides and use the fact that $sqrt(cdot)$ is continuous at zero (note that we approach only from the right) to conclude that
$$
beginaligned
lim_x to 0fracsqrt1 - cos(x^2)1 - cos(x)
&= lim_x to 0sqrtfrac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrtlim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrt2
endaligned$$
as desired.
answered 2 hours ago
Bungo
13.2k22044
By L'Hopital's rule, we have
$$beginaligned
lim_x to 0frac1 - cos(x^2)(1 - cos(x))^2
&= lim_x to 0frac2xsin(x^2)2(1 - cos(x))sin(x) \
&= lim_x to 0fracxsin(x)fracsin(x^2)1 - cos(x) \
&= lim_x to 0fracsin(x^2)1 - cos(x) \
endaligned$$
where we have used that
$$lim_x to 0fracxsin(x) = 1$$
Now we can apply L'Hopital's rule again to obtain
$$beginaligned
lim_x to 0fracsin(x^2)1 - cos(x)
&= lim_x to 0 frac2xcos(x^2)sin(x) \
&= lim_x to 0 fracxsin(x) 2cos(x^2) \
&= lim_x to 0 2cos(x^2) \
&= 2
endaligned$$
Summarizing what we have so far,
$$lim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 = 2$$
Now take the square root of both sides and use the fact that $sqrt(cdot)$ is continuous at zero (note that we approach only from the right) to conclude that
$$
beginaligned
lim_x to 0fracsqrt1 - cos(x^2)1 - cos(x)
&= lim_x to 0sqrtfrac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrtlim_x to 0 frac1 - cos(x^2)(1 - cos(x))^2 \
&= sqrt2
endaligned$$
as desired.
answered 2 hours ago
Bungo
13.2k22044
edited 2 hours ago
answered 2 hours ago
Bungo
13.2k22044
answered 2 hours ago
Bungo
13.2k22044
answered 2 hours ago
Bungo
13.2k22044
13.2k22044
add a comment |Â
add a comment |Â
up vote
1
down vote
if you multiply and divide by both conjugates you get:
$$lim_x to 0 frac(1+cosx)sin(x^2)sqrt1+cos(x^2) space sin^2x$$
Amplifying to get limit of the form $fracsinxx :$
$$=lim_x to 0 frac(1+cosx)sqrt1+cosx^2 space fracsin(x^2)x^2 space fracx^2sin^2x=frac2sqrt2*1*1^2=sqrt2$$
answered 2 hours ago
Villa
44310
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
add a comment |Â
up vote
1
down vote
if you multiply and divide by both conjugates you get:
$$lim_x to 0 frac(1+cosx)sin(x^2)sqrt1+cos(x^2) space sin^2x$$
Amplifying to get limit of the form $fracsinxx :$
$$=lim_x to 0 frac(1+cosx)sqrt1+cosx^2 space fracsin(x^2)x^2 space fracx^2sin^2x=frac2sqrt2*1*1^2=sqrt2$$
answered 2 hours ago
Villa
44310
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
if you multiply and divide by both conjugates you get:
$$lim_x to 0 frac(1+cosx)sin(x^2)sqrt1+cos(x^2) space sin^2x$$
Amplifying to get limit of the form $fracsinxx :$
$$=lim_x to 0 frac(1+cosx)sqrt1+cosx^2 space fracsin(x^2)x^2 space fracx^2sin^2x=frac2sqrt2*1*1^2=sqrt2$$
answered 2 hours ago
Villa
44310
if you multiply and divide by both conjugates you get:
$$lim_x to 0 frac(1+cosx)sin(x^2)sqrt1+cos(x^2) space sin^2x$$
Amplifying to get limit of the form $fracsinxx :$
$$=lim_x to 0 frac(1+cosx)sqrt1+cosx^2 space fracsin(x^2)x^2 space fracx^2sin^2x=frac2sqrt2*1*1^2=sqrt2$$
answered 2 hours ago
Villa
44310
answered 2 hours ago
Villa
44310
answered 2 hours ago
Villa
44310
answered 2 hours ago
Villa
44310
44310
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
add a comment |Â
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
Nice, this is the simplest of the solutions so far. (+1)
– Bungo
2 hours ago
add a comment |Â
up vote
1
down vote
beginalign*
lim_xto 0fracsqrt1-cos(x^2)1-cos(x)
&=lim_xto 0fracsqrt1-cos(x^2)cdot(1+cos x)1-cos^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)sin^2(x)\[10pt]
&=2lim_xto 0fracsqrt1-cos(x^2)x^2cdotfracx^2sin^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)x^2
stackrely:=x^2=2lim_yto 0^+fracsqrt1-cos(y)y\[10pt]
&=2lim_yto 0^+sqrtfrac1-cos(y)2cdotfracsqrt 2y
=2lim_yto 0^2sinleft(frac y2right)cdotfracsqrt 2y
=sqrt 2lim_yto 0^+sinleft(frac y2right)cdotfrac2y=boldsymbolsqrt 2
endalign*
answered 2 hours ago
Pavel R.
25126
add a comment |Â
up vote
1
down vote
beginalign*
lim_xto 0fracsqrt1-cos(x^2)1-cos(x)
&=lim_xto 0fracsqrt1-cos(x^2)cdot(1+cos x)1-cos^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)sin^2(x)\[10pt]
&=2lim_xto 0fracsqrt1-cos(x^2)x^2cdotfracx^2sin^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)x^2
stackrely:=x^2=2lim_yto 0^+fracsqrt1-cos(y)y\[10pt]
&=2lim_yto 0^+sqrtfrac1-cos(y)2cdotfracsqrt 2y
=2lim_yto 0^2sinleft(frac y2right)cdotfracsqrt 2y
=sqrt 2lim_yto 0^+sinleft(frac y2right)cdotfrac2y=boldsymbolsqrt 2
endalign*
answered 2 hours ago
Pavel R.
25126
add a comment |Â
up vote
1
down vote
up vote
1
down vote
beginalign*
lim_xto 0fracsqrt1-cos(x^2)1-cos(x)
&=lim_xto 0fracsqrt1-cos(x^2)cdot(1+cos x)1-cos^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)sin^2(x)\[10pt]
&=2lim_xto 0fracsqrt1-cos(x^2)x^2cdotfracx^2sin^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)x^2
stackrely:=x^2=2lim_yto 0^+fracsqrt1-cos(y)y\[10pt]
&=2lim_yto 0^+sqrtfrac1-cos(y)2cdotfracsqrt 2y
=2lim_yto 0^2sinleft(frac y2right)cdotfracsqrt 2y
=sqrt 2lim_yto 0^+sinleft(frac y2right)cdotfrac2y=boldsymbolsqrt 2
endalign*
answered 2 hours ago
Pavel R.
25126
beginalign*
lim_xto 0fracsqrt1-cos(x^2)1-cos(x)
&=lim_xto 0fracsqrt1-cos(x^2)cdot(1+cos x)1-cos^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)sin^2(x)\[10pt]
&=2lim_xto 0fracsqrt1-cos(x^2)x^2cdotfracx^2sin^2(x)
=2lim_xto 0fracsqrt1-cos(x^2)x^2
stackrely:=x^2=2lim_yto 0^+fracsqrt1-cos(y)y\[10pt]
&=2lim_yto 0^+sqrtfrac1-cos(y)2cdotfracsqrt 2y
=2lim_yto 0^2sinleft(frac y2right)cdotfracsqrt 2y
=sqrt 2lim_yto 0^+sinleft(frac y2right)cdotfrac2y=boldsymbolsqrt 2
endalign*
answered 2 hours ago
Pavel R.
25126
answered 2 hours ago
Pavel R.
25126
answered 2 hours ago
Pavel R.
25126
answered 2 hours ago
Pavel R.
25126
25126
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jkuzmanovik is a new contributor. Be nice, and check out our Code of Conduct.
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You can learn how to write mathematical symbols here.
– Scientifica
2 hours ago
Thanks i was searching for this
– jkuzmanovik
2 hours ago
It's a pleasure :)
– Scientifica
2 hours ago
Does the square root cover only the numerator, or also the denominator?
– Bungo
2 hours ago
Only the numerator
– jkuzmanovik
2 hours ago