# Find lim_{x to oo} {pi/2 - arctan(x)}^(1/x) ?

Jul 28, 2018

${\lim}_{x \to \infty} {\left(\frac{\pi}{2} - \arctan x\right)}^{\frac{1}{x}} = 1$

#### Explanation:

Let $y = \frac{\pi}{2} - \arctan x$.

Then $\cot y = \cot \left(\frac{\pi}{2} - \arctan x\right) = \tan \left(\arctan x\right) = x$ and:

${\lim}_{x \to \infty} y \left(x\right) = 0$

with $y > 0$ bacause $\arctan x < \frac{\pi}{2}$.

So:

${\lim}_{x \to \infty} {\left(\frac{\pi}{2} - \arctan x\right)}^{\frac{1}{x}} = {\lim}_{y \to {0}^{+}} {y}^{\frac{1}{\cot} y}$

As: $\frac{1}{\cot} y = \tan y$:

${\lim}_{x \to \infty} {\left(\frac{\pi}{2} - \arctan x\right)}^{\frac{1}{x}} = {\lim}_{y \to {0}^{+}} {y}^{\tan} y$

Write now the function as:

${y}^{\tan} y = {\left({e}^{\ln} y\right)}^{\tan} y = {e}^{\ln y \tan y}$

and evaluate the limit:

${\lim}_{y \to {0}^{+}} \ln y \tan y$

This is in the indeterminate form $\frac{\infty}{0}$ but we can divide and multiply by $y$ to have:

${\lim}_{y \to {0}^{+}} \left(y \ln y\right) \left(\tan \frac{y}{y}\right)$

and use the well known limits:

${\lim}_{y \to 0} \tan \frac{y}{y} = 1$

${\lim}_{y \to {0}^{+}} y \ln y = 0$

to have:

${\lim}_{y \to {0}^{+}} \ln y \tan y = 0$

and then as ${e}^{t}$ is continuous for every $t \in \mathbb{R}$:

${\lim}_{y \to {0}^{+}} {e}^{\ln y \tan y} = {e}^{\left({\lim}_{y \to {0}^{+}} \ln y \tan y\right)} = {e}^{0} = 1$

In conclusion:

${\lim}_{x \to \infty} {\left(\frac{\pi}{2} - \arctan x\right)}^{\frac{1}{x}} = 1$

graph{(pi/2-arctanx)^(1/x) [-4.75, 15.25, -4.84, 5.16]}