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

Hanul Jeon
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Duy
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3 Answers3

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

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

njguliyev
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  • I seem not too understanding,so, can you explain why"tan(tanx)=..." ? – Duy Oct 06 '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 06 '13 at 13:02
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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 $j=1$.

Did
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  • I seem not too understanding,so, can you explaim why"tan(tanx)=..." ? – Duy Oct 06 '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 06 '13 at 12:20
  • oh, sorry! SO, do you understand "tan(tanx)=..."; and if you know, will you explain for me? – Duy Oct 06 '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 06 '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 08 '17 at 02:45
  • @M.Chen Is your comment actually addressed at me? Then please note that you are in effect asking why $\tan x=x+ax^3+o(x^3)$ for some $a\ne0$. Yes my answer takes this result and similar others for granted (with reasons, I believe, since if you have problems with that, then trying to solve this question becomes preposterous). – Did Aug 08 '17 at 06:50
  • @Did Sorry, Did, I did want to mention you. I think I almost understood your answer after reading it more carefully. I was confused by your neat but sort of strange notation and I thought you had a quick way to find i, j, a, b and n. One more question, did you know the a and b are equal in denominator and numerator respectively by the calculated result or by some straightforward rules or laws? – M. Chen Aug 08 '17 at 10:22
  • @M.Chen What is strange here, one wonders. Anyway, one knows the same factor $(a-b)$ appears in the numerator and in the denominator because the computations in the post apply to every $i$. – Did Aug 08 '17 at 10:25
  • @Did Oh, I think I finally understand your notation f∘i, it means composite f for i times, right? This accounts for some of the 'strange' I said before, by which I also means neat and abstracted. Thank you, Did! – M. Chen Aug 08 '17 at 10:43