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.
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.
Let $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore
$$\lim_{x\to \frac\pi2}\frac{\tan 2x}{x-\frac\pi2}\\ =\lim_{h\to 0}\frac{\tan 2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\ =\lim_{h\to 0}\frac{\tan (\pi+2h)}{h}\\ =\lim_{h\to 0}\frac{\tan 2h}{h}\\ =\lim_{h\to 0}\frac{\sin 2h}{2h}\cdot \frac{2}{\cos 2h}\\ =1\cdot \frac{2}{1}=2$$
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.
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 $2\left(\frac{\pi}{2}-x\right)=t\implies t\to 0 \ as \ x\to \frac{\pi}{2}$ $$=2\times \lim_{t\to 0} \left(\frac{\tan (t)}{(t)}\right)$$ $$=2\times 1=2$$
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!
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 $$\lim_{y\to0}\frac{\tan y}{y}=1$$
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 $u=2x-\pi$.
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.
you can rewrite it as follows:
$$\lim_{x\to\frac\pi2}\frac{\tan2x}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{-\tan(\pi-2x)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan(2x-\pi)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan2(x-\frac\pi2)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{2\tan{(x-\frac\pi2)}}{\Big(1-\tan^2(x-\frac\pi2)\Big)(x-\frac\pi2)}=\lim_{x\to\frac\pi2}\frac{2\tan(x-\frac\pi2)}{x-\frac\pi2}\times \lim_{x\to\frac\pi2}\frac{1}{1-\tan^2(x-\frac\pi2)}=2.1=2$$
$$\because \lim_{x\to 0} \frac{\tan x}{x}=1$$
Shortly, with $t:=2(x-\frac\pi2)$,
$$\lim_{x\to\frac\pi2}\frac{\tan(2x)}{x-\frac\pi2}=2\lim_{t\to0}\frac{\tan(t)}t=2\lim_{t\to0}\frac{\sin(t)}t\lim_{t\to0}\frac1{\cos(t)}.$$
et $x=\frac\pi2 + h$
then as $x\to \frac\pi2$ then $h\to 0$
Therefore
$$\lim_{x\to \frac\pi2}\frac{tan2x}{x-\frac\pi2}\\ =\lim_{h\to 0}\frac{tan2(\frac\pi2+h)}{\frac\pi2+h-\frac\pi2}\\ =\lim_{h\to 0}\frac{tan(\pi+2h)}{h}\\ =\lim_{h\to 0}\frac{tan2h}{h}\\ =\lim_{h\to 0}\frac{(2h) + (2h)^3/3 + ....}{h}\\ =2$$