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Evaluate $$\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}$$

First I tried using L'Hopital's rule..but it's very lengthy

Next I have written the limits as $$L=\lim_{x \to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}=\frac{\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}}{\lim_{x \to 0}\frac{\tan x-\sin x}{x^3}}=\frac{L_1}{L_2}$$

Now by L'Hopital's Rule we get $L_2=0.5$

$$L_1=\lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{x^3}$$

Now $L_1$ can also be evaluated using three applications of L'Hopital's Rule, but is there any other approach?

Martin Sleziak
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Umesh shankar
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1 Answers1

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\begin{align} \lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&= \lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)\cos(\sin x)}{\tan x - \sin x}\\ &=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)+\tan(\sin x)-\tan(\sin x)\cos(\sin x)}{\tan x -\sin x}\\ &=\lim_{x\to 0}\left(\frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}+\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\right) \end{align} Then, by mean value theorem, there is $c\in (-\frac{\pi}{2},\frac{\pi}{2})$ such that $$ \frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\sec^2 c $$ and between $\sin x$ and $\tan x$. Then $\lim_{x\to 0} \frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}=1$. Next, we will show that $\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}=1$. \begin{align} \lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}&=\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))\cos x}{\sin x(1 - \cos x)}\\ &=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\cdot \frac{1-\cos(\sin x)}{\sin^2 x}\cdot \frac{\sin^2 x}{1-\cos x}\cdot \cos x\\ &=\lim_{x\to 0}\frac{\tan(\sin x)}{\sin x}\lim_{x\to 0}\frac{1-\cos(\sin x)}{\sin^2 x}\lim_{x\to 0}(1+\cos x)\lim_{x\to 0}\cos x\\ &=1\cdot \frac{1}{2}\cdot 2\cdot 1\\ &=1. \end{align} Therefore, \begin{align} \lim_{x\to 0}\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x}&=\lim_{x\to 0}\frac{\tan(\tan x)-\tan(\sin x)}{\tan x - \sin x}+\lim_{x\to 0}\frac{\tan(\sin x)(1-\cos(\sin x))}{\tan x - \sin x}\\ &=1+1\\ &=2 \end{align}

choco_addicted
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    Very good solution with elementary technique. Just wanted to add that it is easy to avoid the use of Mean Value Theorem. We can use the formula $\tan a - \tan b = \dfrac{\sin(a - b)}{\cos a\cos b}$ and thus our function is reduced to $$\frac{\sin(\tan x - \sin x)}{\tan x - \sin x}\cdot\frac{1}{\cos(\tan x)\cos(\sin x)} \to 1$$ My upvote +1 – Paramanand Singh Feb 17 '16 at 09:00