Convergence of $\int_0^{\frac{\pi}{2}} \frac{\exp({-1/x)}}{\sqrt{ \sin x}}\, \mathrm{d}x$ 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.
 A: 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.
A: 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{\sin(x)}{e^{-\frac{2}{x}}}
= \lim_{x \searrow 0} \frac{ \frac{d}{d x}\sin(x)}{\frac{d}{d x} e^{-\frac{2}{x}}}
= \lim_{x \searrow 0} \frac{\cos(x)}{\frac{2}{x^2} e^{-\frac{2}{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 one is finite we have by the product rule
\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 $I = 0$.
