# How do you calculate this limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}$?

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

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Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. –  Zarrax Aug 4 '11 at 14:51
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 4 '11 at 14:56
Thanks Zarrax, is just the trick I needed. : D –  mathsalomon Aug 4 '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

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 $1 \cdot 1 = 1$.

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Yess!! thank you very much :D –  mathsalomon Aug 4 '11 at 14:58
@J.J Zarrax might have a bone to pick with you. –  Pedro Tamaroff Feb 22 '12 at 2:32
It might be worth noting that while the solution is pretty natural and standard, in this case you are actually calculating the derivative of $\sin(\sin(x))$ at $x=0$ by using the chain rule. –  N. S. Aug 27 '12 at 16:57

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 $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$

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Taking derivatives are not allowed :( –  user9413 Aug 4 '11 at 15:02
@Chandru Oops. I didn't see that. Thanks. –  Quinn Culver Aug 4 '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 9 '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 0:10

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 4 '11 at 14:57
@mathsalomon: You didn't mention that before :) –  user9413 Aug 4 '11 at 14:58

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 $$\lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1$$

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