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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.

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    @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.2011-08-31

4 Answers 4

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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 $1 \cdot 1 = 1$.

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    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.2012-08-27
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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 $\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$

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    "L'Hô$p$ital's rule makes the $p$roblem much easier": well, I don't thin$k$ it is any simpler than J.J.'s solution.2014-10-22
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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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    @mathsalomon: You didn't mention that before :)2011-08-04
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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 $