Question

${lim}_{x\to \frac{\pi }{2}}\frac{tan2x}{x=\frac{\pi }{2}}$

Answer 1

${lim}_{x\to \frac{\pi }{2}}\frac{tan2x}{x=\frac{\pi }{2}}$
At $x=\frac{\pi }{2}$, the value of the given function takes the form $\frac{0}{0}$
Now, put $x=\frac{\pi }{2}=ysothatx\to \frac{\pi }{2},y\to 0$
$\therefore {lim}_{x\to \frac{\pi }{2}}\frac{tan2x}{x-\frac{\pi }{2}}={lim}_{x\to 0}\frac{tan2(y+\frac{\pi }{2})}{y}$
$={lim}_{x\to 0}\frac{tan(\pi +2y)}{y}$
$={lim}_{y\to 0}\frac{sin2y}{cos2y}$
$=li{m}_{y\to 0}(\frac{sin2y}{2y}\times \frac{2}{cos2y})$
$=\left(li{m}_{2y\to 0}\frac{sin2y}{2y}\right)\times li{m}_{y\to 0}\left(\frac{2}{cosy}\right)[y\to 0\Rightarrow 2y\to 0]$
$=1\times \frac{2}{cos0}[li{m}_{x\to 0}\frac{sinx}{x}=1]$
$=1\times \frac{2}{1}$
$=2$