Question

Evaluate

${lim}_{x\to \frac{\pi }{2}}\frac{\tan 2x}{x-\frac{\pi }{2}}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\frac{\tan 2x}{x-\frac{\pi }{2}}$

let $y=x-\frac{\pi }{2},$ when $x\to \frac{\pi }{2}$

$y\to \frac{\pi }{2}-\frac{\pi }{2}=0$

so, $y\to 0$

Now

$\underset{y\to 0}{lim}\left(\frac{tan2(\frac{\pi }{2}+y)}{y}\right)$

$\Rightarrow \underset{y\to 0}{lim}\left(\frac{tan(\pi +2y)}{y}\right)=\underset{y\to 0}{lim}\left(\frac{tan2y}{y}\right)$

$\Rightarrow \underset{y\to 0}{lim}(\frac{1}{y}.\frac{sin2y}{cos2y})=\underset{y\to 0}{lim}(\frac{sin2y}{y}.\frac{1}{cos2y})$

$=\underset{y\to 0}{lim}\left(\frac{sin2y}{y}\right)\times \underset{y\to 0}{lim}\left(\frac{1}{cos2y}\right)$

$=\underset{y\to 0}{lim}(\frac{sin2y}{y}\times \frac{2y}{2y})\times \underset{y\to 0}{lim}\left(\frac{1}{cos2y}\right)$

$=\underset{y\to 0}{lim}(\frac{sin2y}{y}\times \frac{2y}{2y})\times \underset{y\to 0}{lim}\left(\frac{1}{cos2y}\right)$

$=2\underset{y\to 0}{lim}\left(\frac{sin2y}{2y}\right)\times \underset{y\to 0}{lim}\left(\frac{1}{cos2y}\right)$

$\Rightarrow 2\times 1.\frac{1}{cos\left(0\right)}=\frac{2}{cos0}=\frac{2}{1}=2$

$\therefore \underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\frac{\pi }{2}}=2$