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.
 A: 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.
A: 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 $2\left(\frac{\pi}{2}-x\right)=t\implies t\to 0 \ as \ x\to \frac{\pi}{2}$ $$=2\times \lim_{t\to 0} \left(\frac{\tan (t)}{(t)}\right)$$ $$=2\times 1=2$$
A: And here are two more methods.
METHOD 1:  Exploiting $\lim_{x\to 0}\frac{\sin x}{x}=1$
$$\begin{align}
\lim_{x\to \pi/2}\frac{\tan 2x}{x-\pi/2}&=\lim_{x\to \pi/2}\frac{2\sin x\cos x}{\cos 2x\,(x-\pi/2)}\\\\
&=\lim_{x\to \pi/2}\frac{2\sin x}{\cos 2x}\frac{\cos x}{x-\pi/2}\\\\
&=\lim_{x\to \pi/2}\frac{2\sin x}{\cos 2x}\frac{-\sin( x-\pi/2)}{x-\pi/2}\\\\
&=\left(\lim_{x\to \pi/2}\frac{2\sin x}{\cos 2x}\right)\left(\lim_{x\to \pi/2}\frac{-\sin( x-\pi/2)}{x-\pi/2}\right)\\\\
&=(-2)\,(-1)\\\\
&=2
\end{align}$$

METHOD 2: Asymptotic Approach
Recall that 
$$\tan 2x=2(x-\pi/2)+O((x-\pi/2)^3)$$
Thus
$$\begin{align}
\lim_{x\to \pi/2}\frac{\tan 2x}{x-\pi/2}&=\lim_{x\to \pi/2}\frac{2(x-\pi/2)+O((x-\pi/2)^3)}{x-\pi/2}\\\\
&=\lim_{x\to \pi/2}\left(2+O((x-\pi/2)^2)\right)\\\\
&=2
\end{align}$$
as expected!
A: Let $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore 
$$\lim_{x\to \frac\pi2}\frac{\tan 2x}{x-\frac\pi2}\\
 =\lim_{h\to 0}\frac{\tan 2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\
 =\lim_{h\to 0}\frac{\tan (\pi+2h)}{h}\\
 =\lim_{h\to 0}\frac{\tan 2h}{h}\\
 =\lim_{h\to 0}\frac{\sin 2h}{2h}\cdot \frac{2}{\cos 2h}\\
 =1\cdot \frac{2}{1}=2$$
A: Put $t=x-\pi/2$, we can rewrite limit as
$$\lim_{t\to0}\frac{\tan2 t}{t}=2\lim_{t\to0}\frac{\tan2 t}{2t}=2\cdot1=2$$
by well-known limit
$$\lim_{y\to0}\frac{\tan y}{y}=1$$
A: In this answer, it is shown that
$$
\lim_{x\to0}\frac{\tan(x)}x=1
$$
Therefore, since $\tan(x)=\tan(x-\pi)$, we have
$$
\begin{align}
\lim_{x\to\frac\pi2}\frac{\tan(2x)}{x-\frac\pi2}
&=\lim_{x\to\frac\pi2}2\frac{\tan(2x-\pi)}{2x-\pi}\\
&=2\lim_{u\to0}\frac{\tan(u)}u\\[9pt]
&=2
\end{align}
$$
where $u=2x-\pi$.
A: Hint
Let us consider $$A=\frac{\tan2x}{x-\frac\pi2}$$ For simplicity, change variable $x=y+\frac \pi 2$ which makes $$A=\frac{\tan(2y+\pi)}{y}=\frac{\tan(2y)}{y}=2\frac{\tan(2y)}{2y}$$ You know that, when $z$ is small $\tan(z)\approx z$. Replace $z$ by $2y$ and conclude.
I am sure that you can take from here.
A: you can rewrite it as follows:
$$\lim_{x\to\frac\pi2}\frac{\tan2x}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{-\tan(\pi-2x)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan(2x-\pi)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan2(x-\frac\pi2)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{2\tan{(x-\frac\pi2)}}{\Big(1-\tan^2(x-\frac\pi2)\Big)(x-\frac\pi2)}=\lim_{x\to\frac\pi2}\frac{2\tan(x-\frac\pi2)}{x-\frac\pi2}\times \lim_{x\to\frac\pi2}\frac{1}{1-\tan^2(x-\frac\pi2)}=2.1=2$$ 
$$\because \lim_{x\to 0} \frac{\tan x}{x}=1$$
A: Shortly, with $t:=2(x-\frac\pi2)$,
$$\lim_{x\to\frac\pi2}\frac{\tan(2x)}{x-\frac\pi2}=2\lim_{t\to0}\frac{\tan(t)}t=2\lim_{t\to0}\frac{\sin(t)}t\lim_{t\to0}\frac1{\cos(t)}.$$ 
A: et $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore 
$$\lim_{x\to \frac\pi2}\frac{tan2x}{x-\frac\pi2}\\
 =\lim_{h\to 0}\frac{tan2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\
 =\lim_{h\to 0}\frac{tan(\pi+2h)}{h}\\
 =\lim_{h\to 0}\frac{tan2h}{h}\\
 =\lim_{h\to 0}\frac{(2h) + (2h)^3/3 + ....}{h}\\
 =2$$
