# How do you find the limit of  (x-pi/2)tanx x as it approaches pi/2?

Sep 25, 2015

Use $\cos x = - \sin \left(x - \frac{\pi}{2}\right)$ and the fundamental trigonometric limit.

#### Explanation:

$\left(x - \frac{\pi}{2}\right) \tan x = \left(x - \frac{\pi}{2}\right) \sin \frac{x}{\cos} x$.

Since ${\lim}_{x \rightarrow \frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) = 0$

we should consider trying to use ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = {\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta = 1$ (With $\theta = x - \frac{\pi}{2}$)

To do this we need $\sin \left(x - \frac{\pi}{2}\right)$

Using either the sum formula or using $\cos x = \sin \left(\frac{\pi}{2} - x\right)$ and the fact that sine is an odd function, we get:

$\cos x = - \sin \left(x - \frac{\pi}{2}\right)$.

So,

${\lim}_{x \rightarrow \frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) \tan x = {\lim}_{x \rightarrow \frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) \sin \frac{x}{\cos} x$

 = lim_(xrarr pi/2)(x-pi/2)sinx/(-sin(x-pi/2)

$= {\lim}_{x \rightarrow \frac{\pi}{2}} - \frac{x - \frac{\pi}{2}}{\sin} \left(x - \frac{\pi}{2}\right) \cdot \sin x$

$= - \left(1\right) \left(1\right) = - 1$