Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $1 \cdot 1 = 1$.

share|cite|improve this answer
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 $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$

share|cite|improve this answer
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 :

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

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

share|cite|improve this answer
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 $$ \lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.