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.
$$\lim_{x \to \pi/2} \frac{\log{(\sin x)}}{\frac{1}{\tan x}} = \frac{ \frac{1}{\sin x}( – \cos x )}{\frac{-1}{\sin^2 x}}=\frac{ \frac{1}{1} (0) }{\frac{-1}{1}}=0$$
$$\implies \lim_{x \to \pi/2} (\sin x)^{\tan x} = e^{0}=1$$
Best Answer
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
$$\lim_{x \to \pi/2}\tan x(\sin x-1)=\lim_{x \to \pi/2}\frac{\sin^2x-\sin x}{\cos x}\stackrel{H.R.}=\lim_{x \to \pi/2}\frac{2\sin x\cos x-\cos x}{-\sin x}=0$$