Evaluate $\lim_{x\to 0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}$ 
Solve that $$\lim_{x\to0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}$$

$$
\lim_{x\to 0}\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}=\lim_{x\to 0}\bigg[1+2^{1/\sin^2x}+....+n^{1/\sin^2x}\bigg]^{\sin^2x}\\
[1^t+1^t+....+1^t]^{1/t}=n^{1/t}\leq[1^t+2^t+....+n^t]^{1/t}\leq [n.n^t]^{1/t}=n.n^{1/t}
$$
Can I use squeeze theorem here or is there a better way ?
Note: My reference gives the solution $n$
 A: We have that 
$$n=\bigg[n^{1/\sin^2x}\bigg]^{\sin^2x}\le\bigg[1^{1/\sin^2x}+2^{1/\sin^2x}+…+n^{1/\sin^2x}\bigg]^{\sin^2x}\le \bigg[n\cdot n^{1/\sin^2x}\bigg]^{\sin^2x}\to n$$
A: You were almost there. You have
$$n = (n^t)^{1/t} \le \left(1^t + 2^t + \cdots + n^t \right)^{1/t} \le \left(n \cdot n^t \right)^{1/t} = n \cdot n^{1/t}.$$
Now apply the squeeze theorem using the fact that $n^{1/t} \to 1$ as $t \to \infty$.
A: Let $f(x)$ be the function taken with the limit, then 
\begin{align*}
\log f(x)&=(\sin^{2}x)\log(1^{1/\sin^{2}x}+\cdots+n^{1/\sin^{2}x})\\
&=(\sin^{2}x)\log n^{1/\sin^{2}x}\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{n}{n}\right)^{1/\sin^{2}x}\right)\\
&=(\sin^{2}x)\log n^{1/\sin^{2}x}+(\sin^{2}x)\log\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{n}{n}\right)^{1/\sin^{2}x}\right)\\
&\leq\log n+(\sin^{2}x)\log(1^{1/\sin^{2}x}+\cdots+1^{1/\sin^{2}x})\\
&=\log n+(\sin^{2}x)\log n.
\end{align*}
On the other hand,
\begin{align*}
\log f(x)&\geq\log n+(\sin^{2}x)\log\left(\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}+\cdots+\left(\dfrac{1}{n}\right)^{1/\sin^{2}x}\right)\\
&\geq\log n+(\sin^{2}x)\log\dfrac{1}{n^{(1/\sin^{2}x)-1}}.
\end{align*}
But we know that 
\begin{align*}
(\sin^{2}x)\log\dfrac{1}{n^{(1/\sin^{2}x)-1}}\leq(\sin^{2}x)\cdot\dfrac{1}{n^{(1/\sin^{2}x)-1}}\rightarrow 0\cdot 0=0
\end{align*}
as $x\rightarrow 0$.
So by Squeeze Theorem $\log f(x)\rightarrow\log n$ and hence $f(x)\rightarrow n$.
