Find $\lim_{x\to \frac\pi2}\frac{\tan2x}{x-\frac\pi2}$ without l'hopital's rule.

Solution 1:

Let $x=\frac\pi2 + h$

then as $x\to \frac\pi2$ then $h\to 0$

Therefore

$$\lim_{x\to \frac\pi2}\frac{\tan 2x}{x-\frac\pi2}\\ =\lim_{h\to 0}\frac{\tan 2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\ =\lim_{h\to 0}\frac{\tan (\pi+2h)}{h}\\ =\lim_{h\to 0}\frac{\tan 2h}{h}\\ =\lim_{h\to 0}\frac{\sin 2h}{2h}\cdot \frac{2}{\cos 2h}\\ =1\cdot \frac{2}{1}=2$$

Solution 2:

You might recognize this as the definition of the derivative of $\tan 2x$ at $x = \pi/2$, as this is $$ \lim_{x \to \pi/2} \frac{\tan 2x - \tan \pi}{x - \pi/2},$$ which makes this a very easy derivative exercise.

Solution 3:

Notice $$\lim_{x\to \pi/2} \frac{\tan 2x}{x-\frac{\pi}{2}}$$ $$=\lim_{x\to \pi/2} \frac{-\tan 2\left(\frac{\pi}{2}-x\right)}{x-\frac{\pi}{2}}$$ $$=\lim_{x\to \pi/2} \frac{\tan 2\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)}$$ $$=\lim_{x\to \pi/2} \frac{2\times \tan 2\left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}$$ $$=2\times \lim_{x\to \pi/2} \left(\frac{\tan 2\left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)}\right)$$ Now, let $2\left(\frac{\pi}{2}-x\right)=t\implies t\to 0 \ as \ x\to \frac{\pi}{2}$ $$=2\times \lim_{t\to 0} \left(\frac{\tan (t)}{(t)}\right)$$ $$=2\times 1=2$$