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How do you calculate this limit $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}?$$ without derivatives please. Thanks.

mathsalomon
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    Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. – Zarrax Aug 04 '11 at 14:51
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    Or, intuitively, since $\lim\limits_{x\to 0}\frac{\sin(x)}{x}=1$, then $\sin(x)\approx x$ when $x\approx 0$, so you expect $\sin(\sin(x))\approx \sin(x)\approx x$ when $x$ is very close to $0$. – Arturo Magidin Aug 04 '11 at 14:56
  • Thanks Zarrax, is just the trick I needed. : D – mathsalomon Aug 04 '11 at 14:58
  • @mathsalomon Since a number of nice answers have been given already, please consider accepting one so that the question shows up as answered in the future. – Srivatsan Aug 31 '11 at 12:19

4 Answers4

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Write the limit as $$\lim_{x \to 0} \frac{\sin(\sin x)}{\sin x} \cdot \frac{\sin x}{x}.$$ It is well-known that $$\lim_{x \to 0} \frac{\sin x}{x} = 1,$$ and since $\sin x \to 0$ as $x \to 0$, we get that also $$\lim_{x \to 0} \frac{\sin(\sin x)}{\sin x} = 1.$$ Therefore the limit is $1 \cdot 1 = 1$.

J. J.
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Edit: The solution below should not does not follow the OPs guidelines that derivatives not be used. However, I will leave it since it's correct and shows how L'Hôpital's rule makes the problem much easier. If you think this answer should be deleted, please let me know why and I'll consider it.

Since this limit is of $\frac{0}{0}$ form, we can apply L'Hôpital's rule, which yields $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$

Quinn Culver
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  • Taking derivatives are not allowed :( –  Aug 04 '11 at 15:02
  • @Chandru Oops. I didn't see that. Thanks. – Quinn Culver Aug 04 '11 at 15:05
  • Actually you cannot apply L'H here, because the limit is the definition of the derivative of $\sin( \sin (x))$ at $x=0$. – N. S. Aug 27 '12 at 16:56
  • @N.S. Why does that preclude use of L'H? – Quinn Culver Sep 09 '12 at 21:22
  • Because you USE the derivative of $\sin(\sin(x))$ to calculate ITSELF. That is circular logic.... – N. S. Sep 10 '12 at 00:10
  • @N.S. Why can't I just use the chain rule to conclude $\sin(\sin(x))$ is differentiable at $0$ and then apply L'H? – Quinn Culver Sep 11 '12 at 03:34
  • So, you use chain rule, find that the derivative of $\sin(\sin(x))$ at 0, and then you apply the L'H to calculate again the derivative of $\sin(\sin(x))$ at 0.... – N. S. Sep 11 '12 at 05:31
  • @N.S. It might be redundant, but I don't think it's circular. Moreover, if one didn't notice that the given limit was the definition of $(\sin(\sin x))'$ at $x=0$, then it would be a practical way to proceed. – Quinn Culver Sep 11 '12 at 17:07
  • "L'Hôpital's rule makes the problem much easier": well, I don't think it is any simpler than J.J.'s solution. –  Oct 22 '14 at 07:46
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Note that :

  • $$\sin(\sin{x}) = \sin{x} - \frac{(\sin{x})^{3}}{3!} + \frac{(\sin{x})^{5}}{5!} + \cdots $$

  • $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1$.

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    Thanks Chandru, but I can not use the series expansion when I'm on chapter limits. But thanks for the extraordinary speed in responding. – mathsalomon Aug 04 '11 at 14:57
  • @mathsalomon: You didn't mention that before :) –  Aug 04 '11 at 14:58
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Here is a page with a geometric proof that $$ \lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0}\frac{\tan(x)}{x}=1 $$ You can skip the Corollaries.

Then you can use the fact that $\lim_{x\to 0}\sin(x)=0$ and the fact mentioned by J.J. and Zarrax that $$ \lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1 $$

Argon
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robjohn
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