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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.

muserock92
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  • Your solution is enough. Good job. – Crostul Feb 28 '18 at 14:08
  • Thank you for your reply. I have posted because, in the solution,for $x \in (0,1)$ $f$ is not $0$, but $\frac{4 \log x}{\pi}$ and 0 otherwise, but I can't say why. Thank you. – muserock92 Feb 28 '18 at 14:11
  • @muserock92 in $(0,1) x^{\frac{1}{n}}\to 1; x^n\to 0; \log(x)\to 0$ Thus: $\arctan(x^{\frac{1}{x}})\to \arctan(1)=\frac{\pi}{4}$ –  Feb 28 '18 at 14:13
  • I was thinking something like that, but I can't arrive to that result. P.S.:Are you Italian, like me? Thank you so much. – muserock92 Feb 28 '18 at 14:17
  • @muserock92 Eh già (translation: yes) –  Feb 28 '18 at 14:24

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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):

$\lim_{n\to \infty} \arctan(x^{\frac{1}{n}})+x^n=\arctan{1}+0=\frac{\pi}{4}$