# How do you determine the limit of (x-pi/2)tan(x) as x approaches pi/2?

May 3, 2018

${\lim}_{x \rightarrow \frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) \tan x = - 1$

#### Explanation:

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

$\left(x - \frac{\pi}{2}\right) \tan x$

• $x \to \frac{\pi}{2}$ so $\cos x \ne 0$

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

(xsinx-(πsinx)/2)/cosx

So we need to calculate this limit

lim_(xrarrπ/2)(xsinx-(πsinx)/2)/cosx=_(DLH)^((0/0))

lim_(xrarrπ/2)((xsinx-(πsinx)/2)')/((cosx)' $=$

-lim_(xrarrπ/2)(sinx+xcosx-(πcosx)/2)/sinx $=$

$- 1$

because lim_(xrarrπ/2)sinx=1 ,

lim_(xrarrπ/2)cosx=0

Some graphical help May 3, 2018

For an algebraic solution, please see below.

#### Explanation:

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

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

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

Take limit as $x \rightarrow \frac{\pi}{2}$ using ${\lim}_{t \rightarrow 0} \frac{t}{\sin} t = 1$ to get

${\lim}_{x \rightarrow \frac{\pi}{2}} \left(x - \frac{\pi}{2}\right) \tan x = - 1$