Find $\lim_{x \rightarrow \infty} \frac{\arctan (x^{3/2})}{\sqrt x}$. $$\lim_{x \rightarrow \infty} \frac{\arctan \left(x^\frac{3}{2}\right)}{\sqrt x}.$$
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?
 A: A proper way you are looking for is based on the same reasoning you gave and is called as "squeeze theorem".
Let $g(x)=\frac{\arctan (x^\frac{3}{2})}{\sqrt x}$ then our problem is to find $\displaystyle \lim_{x \to \infty} g(x)$.
Now, as you noticed
\begin{align}
-\frac{\pi}{2}&≤\arctan (x^\frac{3}{2})≤\frac{\pi}{2}\\
\text{or, }-\frac{\pi}{2 \sqrt x} &≤\frac{\arctan (x^\frac{3}{2})}{\sqrt x}≤\frac{\pi}{2 \sqrt x} \end{align} for all $x>0$.
So, there exist two functions $f(x)=-\frac{\pi}{2 \sqrt x},\; h(x)=\frac{\pi}{2 \sqrt x}, \; x \in (0, \infty)$ such that $f(x)≤g(x)≤h(x)$ for all $x>0$
And clearly, $\displaystyle \lim_{x \to \infty} f(x)=\lim_{x \to \infty} h(x)=0$
Therefore, by Squeeze Theorem
$\displaystyle \lim_{x \to \infty} g(x)=0$
A: 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 $f(x)/g(x)\to0$.
