Limit having 2 variables Solve for 
$$\lim_{n\to\infty}\lim_{x\to 0}\frac{1}{(1^{\cot^2x}+2^{\cot^2x}+3^{\cot^2x}+.....+n^{\cot^2x})^{\tan^2x}}$$
Each term in the denominator is of the form of $1^{\infty}$. So
$$\lim_{n\to\infty}\frac{1}{(\lim_{x\to0}1^{\cot^2x}+\lim_{x\to0}2^{\cot^2x}+\lim_{x\to0}3^{\cot^2x}+.....+\lim_{x\to0}n^{\cot^2x})^{\tan^2x}}$$
From where I am not able to find a path to proceed. I'm stuck. Some hints will help me.
 A: Let $\cot^2 x = y\;,$ Then when $x\rightarrow 0\;,$ Then $\cot^2  x = y\rightarrow \infty$
So Limit Convert into $\displaystyle \lim_{y\rightarrow \infty}\frac{1}{\left(1^y+2^y+3^y+....+n^y\right)^{\frac{1}{y}}}$
Now for calculation of  $\displaystyle \lim_{y\rightarrow \infty}\left(1^y+2^y+3^y+....+n^y\right)^{\frac{1}{y}} = n\cdot \lim_{y\rightarrow \infty}\left[\left(\frac{1}{n}\right)^y+\left(\frac{2}{n}\right)^y+\left(\frac{3}{n}\right)^y+..+\left(\frac{n}{n}\right)^y\right]^{\frac{1}{y}} = n$
So Limit  is $\displaystyle = \lim_{n\rightarrow \infty}\frac{1}{n}$
A: The other solution is incomplete. To prove the result properly, one needs to do the following:
$1 \le \left( \sum_{k=1}^n (\frac{k}{n})^y \right)^\frac{1}{y} \le n^\frac{1}{y} \to 1$ as $y \to \infty$.
Therefore by squeeze theorem $\left( \sum_{k=1}^n (\frac{k}{n})^y \right)^\frac{1}{y} \to 1$ as $y \to \infty$.
Note that it is a common mistake to claim $\left( \sum_{k=1}^n (\frac{k}{n})^y \right)^\frac{1}{y} \to \left( \lim_{y\to\infty} \sum_{k=1}^n (\frac{k}{n})^y \right)^\frac{1}{y}$ as $y \to \infty$. The other answer did not do this but many students do.
