$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2 x}}+2^{\frac{1}{\sin^2 x}}+3^{\frac{1}{\sin^2 x}}+\cdots + n^{\frac{1}{\sin^2 x}}\right]^{\sin^2x}$ 
$$\lim_{x\to 0}\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]^{\sin^2x}$$

Limit is  of form $(\infty)^{0} $
$$\lim_{x\to 0}e^{\sin^2x\log{ {\left[1^{\frac{1}{\sin^2x}}+2^{\frac{1}{\sin^2x}}+3^{\frac{1}{\sin^2x}}+\cdots + n^{\frac{1}{\sin^2x}}\right]}}}$$
I don't know how to proceed further.
 A: Hint:
$$n^y\leq 1^y+2^y+3^y+\cdots + n^y\leq n^y+n^y+\cdots+n^y=n^{y+1}$$
A: Note that we can write
$$\begin{align}
\lim_{x\to 0}\left(\sum_{k=1}^n k^{\csc^2(x)}\right)^{\sin^2(x)}&=n\lim_{x\to 0}\left(\sum_{k=1}^n (k/n)^{\csc^2(x)}\right)^{\sin^2(x)} \tag 1\\\\
&=n \tag 2
\end{align}$$
In going from $(1)$ to $(2)$ we proceeded by evaluating the limit
$$\begin{align}
\lim_{x\to 0}\left(\sum_{k=1}^n (k/n)^{\csc^2(x)}\right)^{\sin^2(x)} &=\lim_{x\to 0}e^{\sin^2(x)\log\left(1+\sum_{k=1}^{n-1}(k/n)^{\csc^2(x)}\right)}\\\\
&=\lim_{x\to 0}e^{\sin^2(x)\,O\left(\sum_{k=1}^{n-1}(k/n)^{\csc^2(x)}\right)}\\\\
&=e^0\\\\
&=1 \tag 2
\end{align}$$
A: $$
\begin{align}
&\lim_{x\to0}\left[1^{\frac1{\sin^2(x)}}+2^{\frac1{\sin^2(x)}}+3^{\frac1{\sin^2 (x)}}+\cdots+n^{\frac1{\sin^2(x)}}\right]^{\sin^2(x)}\\
&=\lim_{x\to\infty}\left[1^x+2^x+3^x+\cdots+n^x\right]^{1/x}\\
&=n\lim_{x\to\infty}\left[\left(\frac1n\right)^x+\left(\frac2n\right)^x+\left(\frac3n\right)^x+\cdots+\left(\frac{n-1}n\right)^x+1^x\right]^{1/x}\\[4pt]
&=n\,[0+0+0+\cdots+0+1]^0\\[8pt]
&=n
\end{align}
$$
A: This is not in the form of$(\infty)^0$ as $1^{\infty}$ is undetermined, so try this process.
