# 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.

• Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. Commented Aug 4, 2011 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 $0$. Commented Aug 4, 2011 at 14:56
• Thanks Zarrax, is just the trick I needed. : D Commented Aug 4, 2011 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. Commented Aug 31, 2011 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 $1 \cdot 1 = 1$.

• Yess!! thank you very much :D Commented Aug 4, 2011 at 14:58
• @J.J Zarrax might have a bone to pick with you.
– Pedro
Commented Feb 22, 2012 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. Commented Aug 27, 2012 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 $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$

• Taking derivatives are not allowed :(
– user9413
Commented Aug 4, 2011 at 15:02
• @Chandru Oops. I didn't see that. Thanks. Commented Aug 4, 2011 at 15:05
• Actually you cannot apply L'H here, because the limit is the definition of the derivative of $\sin( \sin (x))$ at $x=0$. Commented Aug 27, 2012 at 16:56
• @N.S. Why does that preclude use of L'H? Commented Sep 9, 2012 at 21:22
• Because you USE the derivative of $\sin(\sin(x))$ to calculate ITSELF. That is circular logic.... Commented Sep 10, 2012 at 0:10

Note that :

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

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

• Thanks Chandru, but I can not use the series expansion when I'm on chapter limits. But thanks for the extraordinary speed in responding. Commented Aug 4, 2011 at 14:57
• @mathsalomon: You didn't mention that before :)
– user9413
Commented Aug 4, 2011 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 $$\lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1$$