TITLE: Find $\lim_{x \rightarrow \infty} \frac{\arctan (x^{3/2})}{\sqrt x}$.
QUESTION [5 upvotes]: $$\lim_{x \rightarrow \infty} \frac{\arctan \left(x^\frac{3}{2}\right)}{\sqrt x}.$$
My method was that as the numerator can never exceed $\pi/2$, some finite value/infinity tends to $0$, but is there a proper way of doing it? As such it's not $\infty/\infty$ or $0/0$  form so l'hopital will not work, so does any other way exist?

REPLY [3 votes]: This argument is definitely an overkill, but it might be worth noting that L'Hopital's rule can be applied if the denominator approaches $\infty$, regardless of whether the numerator does. More precisely, L'Hopital's rule can be applied to a limit of the form $\lim_{x\to\infty}f(x)/g(x)$ if the following conditions are satisfied:

$f$ and $g$ are real-valued functions.
There is an $b\in\mathbb R$ such that $f$ and $g$ are differentiable on $(b,\infty)$.
As $x\to\infty$, $g(x)\to\infty$.
The limit of $f'(x)/g'(x)$ as $x\to\infty$ exists (or equals $\pm\infty$).

For a proof of this fact, see Rudin's Principles of Mathematical Analysis, chapter  5.
If we set $f(x)=\arctan(x^{3/2})$ and $g(x)=\sqrt{x}$, then we see that
$$
\frac{f'(x)}{g'(x)}=\frac{\left(\frac{1}{1+x^3}\right)\frac{3}{2}x^{1/2}}{\frac{1}{2}x^{-1/2}}=\frac{3x}{1+x^3}\to 0 \, .
$$
Hence, $f(x)/g(x)\to0$.
A more elementary argument would be to show that if $f$ is a bounded function, and $g(x)\to\infty$ as $x\to\infty$, then $f(x)/g(x)\to0$.