limit of the sequence of functions $f_n(x) = \frac{\arctan(x^n)}{x^{n+1}}$

Question

As in title, i need to find $$\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n+1}}$$ , where $x \in (0, \infty)$.

When $x \gt 1$, it' easy to see that $\frac{-\pi}{2} \lt \arctan(x^n) \lt \frac{\pi}{2}$, while $x^{n+1} \to \infty$, so $\lim_{n \to \infty}f_n(x) = \infty$.

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

The problem is when $x \in (0, 1)$. My book reads:

Since when $x \in (0, 1)$, $x^{n+1} \to 0$ $$\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n+1}} = \lim_{n \to \infty}\frac{x^n + o(x^n)}{x^{n+1}} = \frac{1}{x}$$

What does it mean? Is this Taylor's theorem in some form? And if it is, how can it be, since it only works when $x \to x_0$, where $x_0$ is its center?

Answer 1

Note that for $x>1$

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

and for $0<x<1$ since $x^n \to 0$ by standard limits

$$\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n+1}}=\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n}}\frac{x^n}{x^{n+1}}=\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n}}\frac{1}{x}=1\cdot \frac1x=\frac1x$$

As a alternative we can use by little-o notation

$$\frac{\arctan(x^n)}{x^{n}} \to 1 \iff \arctan(x^n)=x^n+o(x^n)$$

to conclude as follows

$$\lim_{n \to \infty}\frac{\arctan(x^n)}{x^{n+1}} = \lim_{n \to \infty}\frac{x^n + o(x^{n})}{x^{n+1}} = \lim_{n \to \infty}\left(\frac1x+o\left(\frac1x\right)\right)=\frac{1}{x}$$

since by definition $o\left(\frac1x\right)\to 0$.

Answer 2

They simply use equivalents without saying so: near $0$, $\arctan u\sim u$, so, as in this case $x^n\to 0$, we have $$\frac{\arctan x^n}{x^{n+1}}\sim_{n\to\infty}\frac{x^n}{x^{n+1}}=\frac 1x.$$