Evaluating $\lim_{x \to \pi/2} (\sin x)^{\tan x}$ using Hôpital's Rule I am hoping someone can help me check my work here. I need to evaluate this limit:
$$\lim_{x \to \pi/2} (\sin x)^{\tan x}$$
Since $\sin x$ and $\tan x$ are continuous functions, using the continuity of $e^x$, this expression has the equivalent form:
$$\lim_{x \to \pi/2} e^{\log{((\sin x)^{\tan x})}} $$
$$ \log{(\sin x)^{\tan x}}= \tan x \log{(\sin x)}= \frac{\log{(\sin x)}}{\frac{1}{\tan x}}$$
Taking the limit of this fraction as x goes to $\pi/2$ has the indeterminate form of $-\infty/\infty$, so we can apply Hôpital's rule.
$$\lim_{x \to \pi/2} \frac{\log{(\sin x)}}{\frac{1}{\tan x}} = \frac{ \frac{1}{\sin x}( - \cos x )}{\frac{-1}{\sin^2 x}}=\frac{ \frac{1}{1} (0) }{\frac{-1}{1}}=0$$
$$\implies \lim_{x \to \pi/2} (\sin x)^{\tan x} = e^{0}=1$$
 A: Your way is correct, as an alternative we can use that
$$ (\sin x)^{\tan x}= [(1+(\sin x-1))^{\frac1{\sin x-1}}]^{\tan x(\sin x-1)}\to e^0=1$$
indeed since $t=\sin x-1 \to 0$ by standard limits
$$(1+(\sin x-1))^{\frac1{\sin x-1}}=(1+t)^\frac1t \to e$$
and by l'Hospital
$$\lim_{x \to \pi/2}\tan x(\sin x-1)=\lim_{x \to \pi/2}\frac{\sin^2x-\sin x}{\cos x}\stackrel{H.R.}=\lim_{x \to \pi/2}\frac{2\sin x\cos x-\cos x}{-\sin x}=0$$
A: Setting the limit $\to0$ makes things clearer
$$A=\lim_{x \to \pi/2} (\sin x)^{\tan x}=\lim_{y\to0}(\cos y)^{\cot y}$$
$$\ln A=\lim_{y\to0}\dfrac{\ln(\cos y)}{\tan y}$$  which is of the form $\dfrac00$
$$\ln A=\lim_{y\to0}\dfrac{-\sin y}{\cos y\sec^2y}=0$$
A: Just to give a somewhat different approach, note first that since $\sin x\gt0$ for $x$ near $\pi/2$, we have $\sin x=\sqrt{\tan^2x/(\tan^2x+1)}$, hence
$$(\sin x)^{\tan x}=\left(1+{1\over\tan^2x}\right)^{-(1/2)\tan x}=\left(\left(1+{1\over\tan^2x}\right)^{\tan^2x}\right)^{-1/(2\tan x)}$$
Next, note that for any $u\gt0$ we have
$${1\over e^2}\lt1\lt\left(1+{1\over u}\right)^u\lt e\lt e^2$$
(where the simplifying role of the $e^2$'s will soon be apparent). It follows that
$$\left(1\over e\right)^{1/|\tan x|}\lt(\sin x)^{\tan x}\lt e^{1/|\tan x|}$$
Finally, since $e^{1/|\tan x|}\to1$ as $|\tan x|\to\infty$, the Squeeze Theorem tells us
$$\lim_{x\to\pi/2}(\sin x)^{\tan x}=1$$
