Mixing
Convergence of $int_0^{frac{pi}{2}} frac{exp({-1/x)}}{sqrt{ sin x}}, mathrm{d}x$
$begingroup$
Let $$f(x)=frac{exp(-1/x)}{sqrt{ sin x}}, mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=frac {pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $lim_{xto0+}f(x)$.But I am stuck here as I can't apply Taylor expansions to $e^{-frac{1}{x}}$ and using l'Hospital's rule gives me nothing.
calculus limits improper-integrals
edited Jan 12 at 21:17
Clayton
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
$endgroup$
add a comment |
$begingroup$
Let $$f(x)=frac{exp(-1/x)}{sqrt{ sin x}}, mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=frac {pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $lim_{xto0+}f(x)$.But I am stuck here as I can't apply Taylor expansions to $e^{-frac{1}{x}}$ and using l'Hospital's rule gives me nothing.
calculus limits improper-integrals
edited Jan 12 at 21:17
Clayton
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
$endgroup$
1
$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31
add a comment |
$begingroup$
Let $$f(x)=frac{exp(-1/x)}{sqrt{ sin x}}, mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=frac {pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $lim_{xto0+}f(x)$.But I am stuck here as I can't apply Taylor expansions to $e^{-frac{1}{x}}$ and using l'Hospital's rule gives me nothing.
calculus limits improper-integrals
edited Jan 12 at 21:17
Clayton
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
$endgroup$
Let $$f(x)=frac{exp(-1/x)}{sqrt{ sin x}}, mathrm{d}x.$$
$f(x)$ seems to not have any problems at $x=frac {pi}{2}$, but at $x=0$.
So I should understand behavior of $f(x)$ at $x=0$ by evaluating $lim_{xto0+}f(x)$.But I am stuck here as I can't apply Taylor expansions to $e^{-frac{1}{x}}$ and using l'Hospital's rule gives me nothing.
calculus limits improper-integrals
calculus limits improper-integrals
edited Jan 12 at 21:17
Clayton
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
edited Jan 12 at 21:17
Clayton
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
edited Jan 12 at 21:17
Clayton
19.1k33285
edited Jan 12 at 21:17
Clayton
19.1k33285
edited Jan 12 at 21:17
Clayton
19.1k33285
19.1k33285
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
asked Jan 12 at 21:14
Turan NəsibliTuran Nəsibli
776
776
1
$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31
add a comment |
1
$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31
1
1
$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $e^{1/x}ge 1+frac1x$ for $x>0$ and $xcos(x)le sin(x)$ for $xin [0,pi/2]$.
Hence for $xin(0,pi/2]$, we have
$$0le frac{e^{-1/x}}{sqrt{sin(x)}}le frac{1}{left(1+frac 1xright) sqrt{xcos(x)}}=frac{sqrt x}{x+1}sqrt{frac{
1}{cos(x)}} $$
Now apply the squeeze theorem.
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
|
show 1 more comment
$begingroup$
Using the continuity of the square root, we can conclude
begin{equation*}
I
:=lim_{x searrow 0} frac{e^{-frac{1}{x}}}{sqrt{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{e^{-frac{2}{x}}}{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{1}{e^{frac{2}{x}} sin(x)}}
end{equation*}
Now, we can apply L'Hôpital to the following limit
begin{equation*}
L
:= lim_{x searrow 0} e^{frac{2}{x}} sin(x)
= lim_{x searrow 0} frac{x^2}{2} cos(x) cdot e^{frac{2}{x}}
end{equation*}
And now, because the limits of the factors exist and are finite
begin{equation*}
L = frac{1}{2} underbrace{lim_{x searrow 0} cos(x)}_{= 1} cdot left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)
= frac{1}{2} underbrace{left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)}_{:= widetilde{L}}
end{equation*}
Since $x mapsto x^2$ is monotone and continuous, we have
begin{equation*}
widetilde{L}
= big( underbrace{lim_{x searrow 0} x cdot e^{frac{1}{x}}}_{=: widehat{L}} big)^2
end{equation*}
Now, by L'Hôpital, we have
begin{equation*}
widehat{L}
= lim_{x searrow 0} frac{e^{frac{1}{x}}}{frac{1}{x}}
= lim_{x searrow 0} frac{frac{d}{dx} e^{frac{1}{x}}}{frac{d}{dx} frac{1}{x}}
= lim_{x searrow 0} frac{-frac{1}{x^2} e^{frac{1}{x}}}{-frac{1}{x^2}}
= lim_{x searrow 0} e^{frac{1}{x}}
= infty.
end{equation*}
Therefore, we have $L = infty$ and so $I = 0$.
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
$endgroup$
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $e^{1/x}ge 1+frac1x$ for $x>0$ and $xcos(x)le sin(x)$ for $xin [0,pi/2]$.
Hence for $xin(0,pi/2]$, we have
$$0le frac{e^{-1/x}}{sqrt{sin(x)}}le frac{1}{left(1+frac 1xright) sqrt{xcos(x)}}=frac{sqrt x}{x+1}sqrt{frac{
1}{cos(x)}} $$
Now apply the squeeze theorem.
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
|
show 1 more comment
$begingroup$
Note that $e^{1/x}ge 1+frac1x$ for $x>0$ and $xcos(x)le sin(x)$ for $xin [0,pi/2]$.
Hence for $xin(0,pi/2]$, we have
$$0le frac{e^{-1/x}}{sqrt{sin(x)}}le frac{1}{left(1+frac 1xright) sqrt{xcos(x)}}=frac{sqrt x}{x+1}sqrt{frac{
1}{cos(x)}} $$
Now apply the squeeze theorem.
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
|
show 1 more comment
$begingroup$
Note that $e^{1/x}ge 1+frac1x$ for $x>0$ and $xcos(x)le sin(x)$ for $xin [0,pi/2]$.
Hence for $xin(0,pi/2]$, we have
$$0le frac{e^{-1/x}}{sqrt{sin(x)}}le frac{1}{left(1+frac 1xright) sqrt{xcos(x)}}=frac{sqrt x}{x+1}sqrt{frac{
1}{cos(x)}} $$
Now apply the squeeze theorem.
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
$endgroup$
Note that $e^{1/x}ge 1+frac1x$ for $x>0$ and $xcos(x)le sin(x)$ for $xin [0,pi/2]$.
Hence for $xin(0,pi/2]$, we have
$$0le frac{e^{-1/x}}{sqrt{sin(x)}}le frac{1}{left(1+frac 1xright) sqrt{xcos(x)}}=frac{sqrt x}{x+1}sqrt{frac{
1}{cos(x)}} $$
Now apply the squeeze theorem.
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
edited Jan 13 at 19:16
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
answered Jan 12 at 22:14
Mark ViolaMark Viola
132k1275173
132k1275173
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
|
show 1 more comment
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
Can you please explain your comment on my answer, since your comment is not constructive?
$endgroup$
– Viktor Glombik
Jan 13 at 20:53
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
@viktorglombik For example, your use of LHR after you wrote "Now we can apply ..." makes no sense whatsoever.
$endgroup$
– Mark Viola
Jan 13 at 23:13
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
You can apply LHR if the limit of the quotient of derivatives exists. Have you shown this here? The limit does exist and is $0$, but you haven't shown this. At best your development is convoluted. And writing $L=2(infty)^2$ is definitely nonsense.
$endgroup$
– Mark Viola
Jan 13 at 23:37
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
I have shown $f(x) to infty$, so $frac{1}{f(x)} to 0$ follows, when minding the appropriate restrictions. If you have a good idea on improving my answer, please edit it! We're all here to learn, after all.
$endgroup$
– Viktor Glombik
Jan 14 at 0:57
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
$begingroup$
Viktor I don't have time to rewrite your post.
$endgroup$
– Mark Viola
Jan 14 at 1:24
|
show 1 more comment
$begingroup$
Using the continuity of the square root, we can conclude
begin{equation*}
I
:=lim_{x searrow 0} frac{e^{-frac{1}{x}}}{sqrt{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{e^{-frac{2}{x}}}{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{1}{e^{frac{2}{x}} sin(x)}}
end{equation*}
Now, we can apply L'Hôpital to the following limit
begin{equation*}
L
:= lim_{x searrow 0} e^{frac{2}{x}} sin(x)
= lim_{x searrow 0} frac{x^2}{2} cos(x) cdot e^{frac{2}{x}}
end{equation*}
And now, because the limits of the factors exist and are finite
begin{equation*}
L = frac{1}{2} underbrace{lim_{x searrow 0} cos(x)}_{= 1} cdot left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)
= frac{1}{2} underbrace{left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)}_{:= widetilde{L}}
end{equation*}
Since $x mapsto x^2$ is monotone and continuous, we have
begin{equation*}
widetilde{L}
= big( underbrace{lim_{x searrow 0} x cdot e^{frac{1}{x}}}_{=: widehat{L}} big)^2
end{equation*}
Now, by L'Hôpital, we have
begin{equation*}
widehat{L}
= lim_{x searrow 0} frac{e^{frac{1}{x}}}{frac{1}{x}}
= lim_{x searrow 0} frac{frac{d}{dx} e^{frac{1}{x}}}{frac{d}{dx} frac{1}{x}}
= lim_{x searrow 0} frac{-frac{1}{x^2} e^{frac{1}{x}}}{-frac{1}{x^2}}
= lim_{x searrow 0} e^{frac{1}{x}}
= infty.
end{equation*}
Therefore, we have $L = infty$ and so $I = 0$.
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
$endgroup$
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
add a comment |
$begingroup$
Using the continuity of the square root, we can conclude
begin{equation*}
I
:=lim_{x searrow 0} frac{e^{-frac{1}{x}}}{sqrt{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{e^{-frac{2}{x}}}{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{1}{e^{frac{2}{x}} sin(x)}}
end{equation*}
Now, we can apply L'Hôpital to the following limit
begin{equation*}
L
:= lim_{x searrow 0} e^{frac{2}{x}} sin(x)
= lim_{x searrow 0} frac{x^2}{2} cos(x) cdot e^{frac{2}{x}}
end{equation*}
And now, because the limits of the factors exist and are finite
begin{equation*}
L = frac{1}{2} underbrace{lim_{x searrow 0} cos(x)}_{= 1} cdot left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)
= frac{1}{2} underbrace{left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)}_{:= widetilde{L}}
end{equation*}
Since $x mapsto x^2$ is monotone and continuous, we have
begin{equation*}
widetilde{L}
= big( underbrace{lim_{x searrow 0} x cdot e^{frac{1}{x}}}_{=: widehat{L}} big)^2
end{equation*}
Now, by L'Hôpital, we have
begin{equation*}
widehat{L}
= lim_{x searrow 0} frac{e^{frac{1}{x}}}{frac{1}{x}}
= lim_{x searrow 0} frac{frac{d}{dx} e^{frac{1}{x}}}{frac{d}{dx} frac{1}{x}}
= lim_{x searrow 0} frac{-frac{1}{x^2} e^{frac{1}{x}}}{-frac{1}{x^2}}
= lim_{x searrow 0} e^{frac{1}{x}}
= infty.
end{equation*}
Therefore, we have $L = infty$ and so $I = 0$.
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
$endgroup$
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
add a comment |
$begingroup$
Using the continuity of the square root, we can conclude
begin{equation*}
I
:=lim_{x searrow 0} frac{e^{-frac{1}{x}}}{sqrt{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{e^{-frac{2}{x}}}{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{1}{e^{frac{2}{x}} sin(x)}}
end{equation*}
Now, we can apply L'Hôpital to the following limit
begin{equation*}
L
:= lim_{x searrow 0} e^{frac{2}{x}} sin(x)
= lim_{x searrow 0} frac{x^2}{2} cos(x) cdot e^{frac{2}{x}}
end{equation*}
And now, because the limits of the factors exist and are finite
begin{equation*}
L = frac{1}{2} underbrace{lim_{x searrow 0} cos(x)}_{= 1} cdot left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)
= frac{1}{2} underbrace{left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)}_{:= widetilde{L}}
end{equation*}
Since $x mapsto x^2$ is monotone and continuous, we have
begin{equation*}
widetilde{L}
= big( underbrace{lim_{x searrow 0} x cdot e^{frac{1}{x}}}_{=: widehat{L}} big)^2
end{equation*}
Now, by L'Hôpital, we have
begin{equation*}
widehat{L}
= lim_{x searrow 0} frac{e^{frac{1}{x}}}{frac{1}{x}}
= lim_{x searrow 0} frac{frac{d}{dx} e^{frac{1}{x}}}{frac{d}{dx} frac{1}{x}}
= lim_{x searrow 0} frac{-frac{1}{x^2} e^{frac{1}{x}}}{-frac{1}{x^2}}
= lim_{x searrow 0} e^{frac{1}{x}}
= infty.
end{equation*}
Therefore, we have $L = infty$ and so $I = 0$.
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
$endgroup$
Using the continuity of the square root, we can conclude
begin{equation*}
I
:=lim_{x searrow 0} frac{e^{-frac{1}{x}}}{sqrt{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{e^{-frac{2}{x}}}{sin(x)}}
= sqrt{ lim_{x searrow 0} frac{1}{e^{frac{2}{x}} sin(x)}}
end{equation*}
Now, we can apply L'Hôpital to the following limit
begin{equation*}
L
:= lim_{x searrow 0} e^{frac{2}{x}} sin(x)
= lim_{x searrow 0} frac{x^2}{2} cos(x) cdot e^{frac{2}{x}}
end{equation*}
And now, because the limits of the factors exist and are finite
begin{equation*}
L = frac{1}{2} underbrace{lim_{x searrow 0} cos(x)}_{= 1} cdot left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)
= frac{1}{2} underbrace{left(lim_{x searrow 0} x^2 cdot e^{frac{2}{x}}right)}_{:= widetilde{L}}
end{equation*}
Since $x mapsto x^2$ is monotone and continuous, we have
begin{equation*}
widetilde{L}
= big( underbrace{lim_{x searrow 0} x cdot e^{frac{1}{x}}}_{=: widehat{L}} big)^2
end{equation*}
Now, by L'Hôpital, we have
begin{equation*}
widehat{L}
= lim_{x searrow 0} frac{e^{frac{1}{x}}}{frac{1}{x}}
= lim_{x searrow 0} frac{frac{d}{dx} e^{frac{1}{x}}}{frac{d}{dx} frac{1}{x}}
= lim_{x searrow 0} frac{-frac{1}{x^2} e^{frac{1}{x}}}{-frac{1}{x^2}}
= lim_{x searrow 0} e^{frac{1}{x}}
= infty.
end{equation*}
Therefore, we have $L = infty$ and so $I = 0$.
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
edited Jan 14 at 0:56
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
answered Jan 12 at 21:57
Viktor GlombikViktor Glombik
9251527
9251527
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
add a comment |
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
2
2
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
$begingroup$
This makes no sense.
$endgroup$
– Mark Viola
Jan 12 at 22:09
add a comment |
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$begingroup$
You could try squaring $f$ first, then compute the limit at $0$.
$endgroup$
– user170231
Jan 12 at 21:25
$begingroup$
@user170231 what exactly lets us squaring $f$ and then take the limit?
$endgroup$
– Turan Nəsibli
Jan 12 at 21:31