Find $\lim_{x\to \frac\pi2}\frac{\tan2x}{x-\frac\pi2}$ without l'hopital's rule. I'm required to find $$\lim_{x\to\frac\pi2}\frac{\tan2x}{x-\frac\pi2}$$ without l'hopital's rule.
Identity of $\tan2x$ has not worked. 
Kindly help.

et $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore 
$$\lim_{x\to \frac\pi2}\frac{tan2x}{x-\frac\pi2}\\
 =\lim_{h\to 0}\frac{tan2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\
 =\lim_{h\to 0}\frac{tan(\pi+2h)}{h}\\
 =\lim_{h\to 0}\frac{tan2h}{h}\\
 =\lim_{h\to 0}\frac{(2h) + (2h)^3/3 + ....}{h}\\
 =2$$
