Question

Evaluate: $\underset{x\to \frac{\pi }{2}}{lim}(2x\tan x-\frac{\pi }{\cos x})$

Answer 1

$\underset{x\to \frac{\pi }{2}}{lim}(2x\tan x-\frac{\pi }{\cos x})$

$=\underset{x\to \frac{\pi }{2}}{lim}(\frac{2x\sin x-\pi }{\cos x})$

L'Hopital's rule for indeterminate forms:

In case of an indeterminate form of the type

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}$

or

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\infty }{\infty }$

where $a$ is any real number, positive infinity or negative infinity, then

$\underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Therefore, $\underset{x\to \frac{\pi }{2}}{lim}(\frac{2x\sin x-\pi }{\cos x})$

$=\underset{x\to \frac{\pi }{2}}{lim}(\frac{2\sin x+2x\cos x}{-\sin x})$

$=\frac{2+0}{-1}$

$=-2$