9
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What's with all of the questions about computing limits without using l'Hopital's rule? Is this a standard thing in calculus classes? $\endgroup$ – anomaly Jul 16 '15 at 4:59
  • 2
    $\begingroup$ @anomaly Sir I'm a class XII student and in book l'Hopital's rule is not given . So if I do the problem correctly by using l'Hopital's rule then I'll get 0 marks on that. So it is important for me to learn solving these kind of problems without the l'Hopital's rule. Sir you must understand the condition of a student specially in school. $\endgroup$ – golu Jul 16 '15 at 5:05
  • 3
    $\begingroup$ I'm asking why classes assign such questions, not whether you should use it on this homework assignment. L'Hopital's rule is trivial to prove and a great tool for computing various limits. If you want to test students' ability to compute various limits without resorting to l'Hopital's rule, ask them questions where l'Hopital's rule doesn't instantly give you the answer, rather than artificially tying students' hands behind their backs. $\endgroup$ – anomaly Jul 16 '15 at 5:10
  • $\begingroup$ @anomaly: From OP comment it is clear that he is studying calculus as a part of high school and not studying real-analysis in a typical undergrad course. At this stage many students are simply not taught any proofs in calculus. LHR proof is a big big deal for them and certainly not trivial. $\endgroup$ – Paramanand Singh Jul 18 '15 at 12:29
  • $\begingroup$ @ParamanandSingh: If you're not covering proofs, then there's even less reason to avoid l'Hopital's rule. $\endgroup$ – anomaly Jul 18 '15 at 12:50

10 Answers 10

11
$\begingroup$

Let $x=\frac\pi2 + h$

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

Therefore

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

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Isn't that L'Hosipital in disguise ? $\endgroup$ – Yves Daoust Jul 16 '15 at 7:08
  • 2
    $\begingroup$ @YvesDaoust no, mixedmath is saying that this literally follows from the definition of the derivative of a function. $\endgroup$ – hunter Jul 16 '15 at 7:11
  • $\begingroup$ @hunter: I know, but this is essentially L'Hospital. $(f(x)-f(a))/(x-a)\leftrightarrow\frac{f'(a)}1$. $\endgroup$ – Yves Daoust Jul 16 '15 at 7:16
  • 1
    $\begingroup$ @YvesDaoust: I think you are mixing two things. If the limit of $(f(x) - f(a))/(x - a)$ as $x \to a$ exists then by definition this limit is called derivative of $f$ at $a$ and denoted by $f'(a)$. This definition requires that the function be defined in a certain neighborhood of $a$. On the other LHR would require that $f'(x)$ be defined in some neighborhood of $a$ (except possibly at $x = a$). So it is not essentially LHR. One is a definition and another is a significant theorem with not so easy proof. $\endgroup$ – Paramanand Singh Jul 18 '15 at 12:21
  • $\begingroup$ @ParamanandSingh: from an operational point of view, I cannot see a difference. You are computing a derivative, in the symbolic way. $\endgroup$ – Yves Daoust Jul 18 '15 at 12:28
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $$\lim_{y\to0}\frac{\tan y}{y}=1$$

$\endgroup$
1
$\begingroup$

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 $u=2x-\pi$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

you can rewrite it as follows:

$$\lim_{x\to\frac\pi2}\frac{\tan2x}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{-\tan(\pi-2x)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan(2x-\pi)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{\tan2(x-\frac\pi2)}{x-\frac\pi2}=\lim_{x\to\frac\pi2}\frac{2\tan{(x-\frac\pi2)}}{\Big(1-\tan^2(x-\frac\pi2)\Big)(x-\frac\pi2)}=\lim_{x\to\frac\pi2}\frac{2\tan(x-\frac\pi2)}{x-\frac\pi2}\times \lim_{x\to\frac\pi2}\frac{1}{1-\tan^2(x-\frac\pi2)}=2.1=2$$

$$\because \lim_{x\to 0} \frac{\tan x}{x}=1$$

$\endgroup$
0
$\begingroup$

Shortly, with $t:=2(x-\frac\pi2)$,

$$\lim_{x\to\frac\pi2}\frac{\tan(2x)}{x-\frac\pi2}=2\lim_{t\to0}\frac{\tan(t)}t=2\lim_{t\to0}\frac{\sin(t)}t\lim_{t\to0}\frac1{\cos(t)}.$$

$\endgroup$
0
$\begingroup$

et $x=\frac\pi2 + h$

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

Therefore

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.