How to find the limit of $\lim_{n \to +\infty}{n \left(\arctan\frac{1}{\sqrt n}\right)^{2n}}$?

Question

I have literally no idea on how to solve this one $$\lim_{n \to +\infty}{n \left(\arctan\frac{1}{\sqrt n}\right)^{2n}}$$

Answer 1

Notice that

$$\arctan x = \int_0^x\frac{1}{1+t^2}\,dt \le x$$

for all $x\ge 0.$ Thus

$$n\left (\arctan 1/\sqrt n \right)^n \le n\cdot (1/\sqrt n)^n.$$

What do you think?

Answer 2

just a hint

Using the fact that $$\arctan (x)\sim x \;(x\to 0) $$ we have $$u_n\sim n (n^{-\frac12})^{2n} \;\;(n\to +\infty) $$

$$\sim n^{1-n} \;\; (n\to+\infty) $$

but $$n^{1-n}=e^{n(\frac1n-1)\ln (n)} $$

thus, the limit is $e^{-\infty}= $zero.

Answer 3

Consider instead $$ \lim_{x\to0^+}\frac{1}{x^2}(\arctan x)^{2/x^2} $$ The limit of the logarithm is $$ \lim_{x\to0^+} 2\left( \frac{\log\arctan x}{x^2}-\log x \right)= 2\lim_{x\to0^+}\frac{\log\arctan x-x^2\log x}{x^2}=-\infty $$ because $\lim_{x\to0^+}x^2\log x=0$.

Hence your limit is $0$, because $\lim_{t\to-\infty}e^t=0$.