Question

Evaluate: $\underset{x\to \frac{\pi }{2}}{lim}\frac{tan2x}{x-\frac{\pi }{2}}$

Answer 1

We have, $\underset{x\to \frac{\pi }{2}}{lim}\frac{tan2x}{x-\frac{\pi }{2}}$

At $x=\frac{\pi }{2}$, the value of the given function takes the form $\frac{0}{0}$

If, we put $x=h+\frac{\pi }{2}$ such that $x\to \frac{\pi }{2}thenh\to 0$

Therefore $\underset{x\to \frac{\pi }{2}}{lim}\frac{tan2x}{x-\frac{\pi }{2}}=\underset{h\to 0}{lim}\frac{tan2(h+\frac{\pi }{2})}{h}$

$\underset{h\to 0}{lim}\frac{tan(\pi +2h)}{h}\left\{tan\right(\pi +\theta )=tan\theta \}$

$=\underset{h\to 0}{lim}\frac{tan2h}{h}\times \frac{2}{2}$

$=2\underset{h\to 0}{lim}\frac{tan2h}{h}\{\underset{x\to 0}{lim}\frac{tanx}{x}+1\}$

$=2\times 1=2$