Question

Solve the equation:-
${lim}_{x\to \pi /2}\frac{\tan 2x}{x-\pi /2}$

Answer 1

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

$LHL={lim}_{h\to 0^{-}}\frac{\tan 2(\frac{\pi }{2}-h)}{\frac{\pi }{2}-h-\frac{\pi }{2}}$

$={lim}_{h\to 0^{-}}\frac{\tan (\pi -2h)}{-h}$

$={lim}_{h\to 0^{-}}-\frac{\tan 2h}{-h}$

$={lim}_{h\to 0^{-}}-\frac{2\tan 2h}{-2h}=2$

$RHL={lim}_{h\to 0^{+}}\frac{\tan 2(\frac{\pi }{2}+h)}{\frac{\pi }{2}+h-\frac{\pi }{2}}$

$={lim}_{h\to 0^{+}}\frac{\tan (\pi +2h)}{h}$

$={lim}_{h\to 0^{+}}\frac{\tan 2h}{h}$

$={lim}_{h\to 0^{+}}\frac{2\tan 2h}{2h}=2$

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