# Find lim:$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$

Find lim: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}$$. You can use L'Hospitale, or Maclaurin, etc

One way without MacLaurin or L'Hospital: $$\lim_{x\to0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x}=\lim_{x\to0} \frac{\tan(\tan x) - \tan(\sin x)+\tan(\sin x)-\sin(\sin x)}{\tan x -\sin x} = \lim_{x\to0} \frac{[\tan(\tan x -\sin x)][1-\tan(\tan x)\tan(\sin x)]}{\tan x -\sin x}+\lim_{x\to0} \frac{\sin(\sin x)[1 - \cos(\sin x)]}{\cos(\sin x)(\tan x -\sin x)}=$$ $$=1+\lim_{x\to0} \frac{\sin(\sin x)}{\sin x}\lim_{x\to0} \frac{1 - \cos(\sin x)}{\sin^2 x}\lim_{x\to0}\frac{x^2}{1-\cos x}\lim_{x\to0}\frac{\sin^2x}{x^2} = 1+1.\frac{1}{2}.2.1 =2$$.

We used $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b}$$ $$\lim_{x\to0}\frac{\sin x}{x}= \lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\cos x=1, \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$

• very good answer without the use of L'Hospital. +1 – Paramanand Singh Nov 17 '14 at 8:53

Hint: Use $\tan x = x+\dfrac{x^3}{3}+o(x^3)$ and $\sin x = x-\dfrac{x^3}{6}+o(x^3)$ to obtain $\tan(\tan x) = x+\dfrac{2x^3}{3}+o(x^3)$ and $\sin (\sin x) = x-\dfrac{x^3}{3}+o(x^3)$.

Hence $$\dfrac{\tan(\tan x) - \sin(\sin x)}{\tan x -\sin x} = \dfrac{x^3+o(x^3)}{\frac{x^3}{2}+o(x^3)} \to 2.$$

• I seem not too understanding,so, can you explain why"tan(tanx)=..." ? – Duy Oct 6 '13 at 12:17
• $\tan(\tan x) = \tan x + \dfrac{\tan^3 x}{3}+o(\tan^3 x) = x+\dfrac{x^3}{3}+o(x^3) + \dfrac13(x+\dfrac{x^3}{3}+o(x^3))^3 + o(x^3)$. – njguliyev Oct 6 '13 at 13:02

Assume that $h(x)\to0$ when $x\to0$ and that $f$ and $g$ are such that $$f(x)=h(x)+ah(x)^{1+n}+o(h(x)^{1+n}),$$ and $$g(x)=h(x)+bh(x)^{1+n}+o(h(x)^{1+n}),$$ with $a\ne b$, for some $n\gt0$. Then, for every $i\geqslant1$, $$f^{\circ i}(x)=h(x)+iah(x)^{1+n}+o(h(x)^{1+n}),$$ and $$g^{\circ i}(x)=h(x)+ibh(x)^{1+n}+o(h(x)^{1+n}),$$ hence $$f^{\circ i}(x)-g^{\circ i}(x)\sim i(a-b)h(x)^{1+n},$$ and, for every positive $i$ and $j$, $$\lim_{x\to0}\frac{f^{\circ i}(x)-g^{\circ i}(x)}{f^{\circ j}(x)-g^{\circ j}(x)}=\frac{i}j.$$ Application: for $h(x)=x$, $f(x)=\tan x$, $g(x)=\sin x$ hence $n=2$, $a=\frac13\ne-\frac16=b$, choose $i=2$ and $j=1$.

• I seem not too understanding,so, can you explaim why"tan(tanx)=..." ? – Duy Oct 6 '13 at 12:16
• OK. But I am afraid you will have to explain which parts you do not understand (there is no "tan(tan(x)=..." in my post). (Was this comment meant for the other answer?) – Did Oct 6 '13 at 12:20
• oh, sorry! SO, do you understand "tan(tanx)=..."; and if you know, will you explain for me? – Duy Oct 6 '13 at 12:41
• "Understand"? What is there to understand to "tan(tanx)=..." (which, once again, is not in my answer)? Sorry, I do not follow. – Did Oct 6 '13 at 12:43
• @Duy Duy, your answer is interesting, but I don't know how do you get the n, i and j. Could you tell me the name or general idea of your method, so I can read some reference to understand your notation and answer. – M. Chen Aug 8 '17 at 2:45