I'm working with this sequence of functions:
$$f_n(x)=\frac{\log x}{\arctan{x^{1/n}}+x^{n}}$$
for $n\geq 2$ and $x \in (0,+\infty)$.
I have to find $f$ such that $f_{n} \rightarrow f$ pointwise in $(0,+\infty)$.
So, my idea is to compute the $$\lim_{n\to \infty}f_{n}(x) < \infty.$$
Computing such limit:
$$\lim_{n\to \infty}\frac{\log x}{\arctan{x^{1/n}}+x^{n}}=\lim_{n\to \infty}\frac{\log x}{x^{n}}=0.$$
So I can say that $f_{n} \rightarrow f$ pointwise with $f=0$.
Is this sufficient to answer the question? May I have to consider some other cases? Thank you very much.
EDIT I'm aking this because in the solution the answer is not $0$ everywhere but:
$$f= \begin{cases} \frac{4 \log x}{\pi} , & \text{if $x \in (0,1)$} \\ 0, & \text{if $x\geq 1$} \end{cases}$$
But I don't know how to prove that. Thank you.
When $x>1$, you have that $\lim_{n\to \infty}\arctan(x^{\frac{1}{n}})=o(x^n)$
Thus, the denumerator is $O(x^n)$, while the numerator is $o(x^n)$, and so $\lim_{n\to \infty}f_n=0$.
For $0<x\leq1$ the reasoning is different (since the n-th root goes to 1, while the n-th power goes to 0):
$\lim_{n\to \infty} \arctan(x^{\frac{1}{n}})+x^n=\arctan{1}+0=\frac{\pi}{4}$