Question

Evaluate the given limit :
$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan \left(2x\right)}{x-\frac{\pi }{2}}$

Answer 1

Given :$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan \left(2x\right)}{x-\frac{\pi }{2}}$
Substituting $x=\frac{\pi }{2}$ in the given limit,
$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan \left(2x\right)}{x-\frac{\pi }{2}}=\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2\left(\frac{\pi }{2}\right)}{\frac{\pi }{2}-\frac{\pi }{2}}$
$=\frac{\tan \pi }{0-0}$
$=\frac{0}{0}$
Since it is in $\frac{0}{0}$ form.

We need to simplify it,
Let $L=\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\frac{\pi }{2}}$
Let $y=x-\frac{\pi }{2}\Rightarrow y+\frac{\pi }{2}=x$

When $x\to \frac{\pi }{2}$
$y\to \frac{\pi }{2}-\frac{\pi }{2}$
$y\to 0$
So, our equations become
$\Rightarrow L=\underset{y\to 0}{lim}\left(\frac{\tan 2(\frac{\pi }{2}+y)}{y}\right)$
$\Rightarrow L=\underset{y\to 0}{lim}\left(\frac{\tan (\pi +2y)}{y}\right)$
$\Rightarrow L=\underset{y\to 0}{lim}\left(\frac{\tan 2y}{y}\right)$
$\Rightarrow L=\underset{y\to 0}{lim}(\frac{1}{y}\cdot \frac{\sin 2y}{\cos 2y})$
$\Rightarrow L=\underset{y\to 0}{lim}(\frac{\sin 2y}{y}\cdot \frac{1}{\cos 2y})$
$\Rightarrow L=\underset{y\to 0}{lim}\frac{\sin 2y}{y}\times \underset{y\to 0}{lim}\frac{1}{\cos 2y}$
Multiply and divide by $2$
$\Rightarrow L=\underset{y\to 0}{lim}(\frac{\sin 2y}{y}\times \frac{2}{2})\cdot \underset{y\to 0}{lim}\frac{1}{\cos 2y}$
$\Rightarrow L=2\underset{y\to 0}{lim}\left(\frac{\sin 2y}{2y}\right)\cdot \underset{y\to 0}{lim}\frac{1}{\cos 2y}$
$\Rightarrow L=2\times 1\times \underset{y\to 0}{lim}\frac{1}{\cos 2\left(0\right)}$
$\Rightarrow L=2\times \frac{1}{1}$
$\Rightarrow L=2$