Question

Prove that:
$\underset{x\to \frac{\pi }{2}}{lim}\frac{\tan 2x}{x-\frac{\pi }{2}}$

Answer 1

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

Put $y=x-\frac{\pi }{2}$

as $x\to \pi /2.$

$y=\frac{\pi }{2}-\frac{\pi }{2.}$

$y\to 0$

so our equation is

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

${lim}_{y\to 0}\frac{tan(\pi +2y)}{y}$

${lim}_{y\to 0}\frac{tan2y}{y}$

${lim}_{y\to 0}(\frac{1}{y}.\frac{sin2y}{cos2y})$

${lim}_{y\to 0}\frac{sin2y}{y}\times {lim}_{y\to 0}\frac{1}{cos2y}$

${lim}_{y\to 0}(\frac{sin2y}{2y}\times \frac{2y}{y}).{lim}_{y\to 0}\frac{1}{cos2y}$

${lim}_{y\to 0}(\frac{sin2y}{2y}\times 2){lim}_{y\to 0}\frac{1}{cos2y}$

$2.{lim}_{y\to 0}\frac{sin2y}{2y}.{lim}_{y\to 0}\frac{1}{cos2y}$

$\mathrm{2.1.}{lim}_{y\to 0}\frac{1}{cos2y}$

$2.\frac{1}{cos2/0}$

$=\frac{2}{cos0}$

$=\frac{2}{1}=2$