Evaluate $ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $ $ \lim_{n\to \infty} ( \lim_{x\to0} (1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} $
The answer should be $ {e}^\frac{1}{3} $
I haven't encountered problems like this before and I'm pretty confused, thank you.
I guess we must use the remarkable limit of $ \frac{\tan(x)}{x} $ when $ x $ approaches $ 0 $ by dividing and then multiplaying with $x$.

Note that for $x\to 0$


*

*$\tan x= x+o(x^2)$


then
$$(1+\tan^2(x)+\tan^2(2x)+ \cdots + \tan^2(nx)))^{\frac{1}{n^3x^2}} = (1+x^2+4x^2+...+n^2x^2+o(x^2)) ^ {\frac{1}{n^3x^2}}=\large{e^{\frac{\log(1+x^2+4x^2+...+n^2x^2+o(x^2))}{n^3x^2}}=e^{\frac{x^2+4x^2+...+n^2x^2+o(x^2)}{n^3x^2}}=e^{\frac{n(2n+1)(n+1)}{6n^3}+o(1)} \stackrel{x\to0}\to e^{\frac{n(2n+1)(n+1)}{6n^3}}}$$
and for $n\to \infty$
$$\Large{e^{\frac{n(2n+1)(n+1)}{6n^3}}}\to e^\frac13$$
