# How do you use the Squeeze Theorem to find lim Arctan(n^2)/sqrt(n) as x approaches infinity?

Oct 18, 2015

See the explanation, below.

#### Explanation:

For all $x$, we know that ${x}^{2} > 0$, so we have

$0 \le \arctan \left({x}^{2}\right) < \frac{\pi}{2}$.

For positive $x$, $\sqrt{x} > 0$, so we can divide the inequalitiy without changing the directions of the inequalities.

$0 \le \arctan \frac{{x}^{2}}{\sqrt{x}} < \frac{\pi}{2 \sqrt{x}}$.

${\lim}_{x \rightarrow \infty} 0 = 0$ and ${\lim}_{x \rightarrow \infty} \frac{\pi}{2 \sqrt{x}} = 0$

Therefore,

${\lim}_{x \rightarrow \infty} \arctan \frac{{x}^{2}}{\sqrt{x}} = 0$