Calculating $\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan x}$ Please help me calculate: $\lim_{x\to\frac{\pi}{2}}(\sin x)^{\tan x}$
 A: Take $y=(\sin x)^{\tan x}$
Taking log on both sides we have,
$\log y=\tan x\log(\sin x)=\frac{\log(\sin x)}{\cot x}$
Now as $x\to \pi/2$, $\log(\sin x)\to 0$ and $\cot x\to 0$
Now you can use L'Hospital's Rule.
$$\lim_{x\to \pi/2}\frac{\log(\sin x)}{\cot x}=\lim_{x\to \pi/2}\frac{\cos x}{\sin x(-\csc^2 x)}=\lim_{x\to \pi/2}\frac{\cos x}{-\sin x}=0$$
$$\Rightarrow \log y\to 0, \text{as}, x\to \pi/2$$
$$\Rightarrow  y\to \exp^0, \text{as}, x\to \pi/2$$
$$\Rightarrow  y\to 1, \text{as}, x\to \pi/2$$
A: HINT
$$\lim_{x \to a}f(x)^{g(x)} = \left(\lim_{x \to a} f(x) \right)^{\lim_{x \to a}g(x)}$$ if $f(x)$ and $g(x)$ are continuous in the neighborhood of $a$ and $\lim_{x \to a} f(x) \in \mathbb{R}, \lim_{x \to a}g(x) \in \mathbb{R}$. This is all you need to compute the limit you want.

This is the answer to the new question.
\begin{align}
\underbrace{\lim_{x \to \pi/2} (\sin(x))^{\tan(x)}  = \lim_{t \to 0} \cos(t)^{\cot(t)}}_{t \mapsto \pi/2-x} &= \lim_{t \to 0} \left(1 - 2 \sin^2(t/2)\right)^{\cot(t)}\\
&= \lim_{t \to 0} \left(1 - 2 \sin^2(t/2)\right)^{\frac1{2\sin^2(t/2)} \cdot 2 \sin^2(t/2)\cot(t)}
\end{align}
Now $$2 \sin^2(t/2)\cot(t) =2\sin^2(t/2) \dfrac{\cos(t)}{\sin(t)} = 2\sin^2(t/2) \dfrac{\cos(t)}{2 \sin(t/2) \cos(t/2)} = \cos(t) \tan(t/2)$$
Now let $f(t) = \left(1 - 2 \sin^2(t/2)\right)^{\frac1{2\sin^2(t/2)}}$, $g(t) = \cos(t) \tan(t/2)$ and $a=0$.
You might also make use of the following limit
$$\lim_{y \to 0}(1-y)^{1/y} = \dfrac1e$$ to evaluate $\lim_{t \to 0} f(t)$.
