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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 $\lim_{x\to0+}f(x)$.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.

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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 Nasibli Jan 12 at 21:31
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Note that $e^{1/x}\ge 1+\frac1x$ for $x>0$ and $x\cos(x)\le \sin(x)$ for $x\in [0,\pi/2]$.

Hence for $x\in(0,\pi/2]$, we have

$$0\le \frac{e^{-1/x}}{\sqrt{\sin(x)}}\le \frac{1}{\left(1+\frac 1x\right) \sqrt{x\cos(x)}}=\frac{\sqrt x}{x+1}\sqrt{\frac{ 1}{\cos(x)}} $$

Now apply the squeeze theorem.

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  • $\begingroup$ 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
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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 $I = 0$.

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    $\begingroup$ This makes no sense. $\endgroup$ – Mark Viola Jan 12 at 22:09

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