Trigonometric Limit tends to $\infty$

Question

$$\lim_{x \to 0}\lim_{n \to \infty} \frac{n}{\left(1^\left(\cot^2x \right)+2^\left(\cot^2x \right)+...+n^\left(\cot^2x \right)\right)^\left(\tan^2x \right)}$$

How to approach? Need hints.

Answer 1

Expanding on my comment, the expression under limit can be expressed as $$\left(\frac{n^{a+1}}{1^{a}+2^{a}+\cdots +n^{a}} \right) ^{1/a}\cdot n^{-1/a}$$ where $a=\cot^{2}x$ and this clearly tends to $(a+1)^{1/a}\cdot 0=0$.

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The question would be more interesting if presented as $$\lim_{x\to 0}\lim_{n\to\infty}\frac{n^{\operatorname{cosec}^{2}x}}{(1^{\tan^{2}x}+2^{\tan^{2}x}+\cdots+n^{\tan^{2}x})^{\cot^{2}x}}$$ The expression under limit operation tends to $(1+b)^{1/b}$ as $n\to\infty$ where $b=\tan^{2}x$. Next as $x\to 0$ the variable $b\to 0$ and therefore $(1+b)^{1/b}\to e$ so that final answer is $e$. Another variation which is more in line with the given question is $$\lim_{x\to 0}\lim_{n\to\infty} \frac{n^{\sec^{2}x}}{(1^{\cot^{2}x}+2^{\cot^{2}x}+\cdots+n^{\cot^{2}x})^{\tan^{2}x}}$$ and the final answer is then $1$.