13
$\begingroup$

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

$\endgroup$
4
  • 23
    $\begingroup$ Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. $\endgroup$
    – Zarrax
    Aug 4 '11 at 14:51
  • 3
    $\begingroup$ 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 $0$. $\endgroup$ Aug 4 '11 at 14:56
  • $\begingroup$ Thanks Zarrax, is just the trick I needed. : D $\endgroup$ Aug 4 '11 at 14:58
  • $\begingroup$ @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. $\endgroup$
    – Srivatsan
    Aug 31 '11 at 12:19
22
$\begingroup$

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 $1 \cdot 1 = 1$.

$\endgroup$
3
  • $\begingroup$ Yess!! thank you very much :D $\endgroup$ Aug 4 '11 at 14:58
  • 3
    $\begingroup$ @J.J Zarrax might have a bone to pick with you. $\endgroup$
    – Pedro Tamaroff
    Feb 22 '12 at 2:32
  • $\begingroup$ 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. $\endgroup$
    – N. S.
    Aug 27 '12 at 16:57
10
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ Taking derivatives are not allowed :( $\endgroup$
    – user9413
    Aug 4 '11 at 15:02
  • $\begingroup$ @Chandru Oops. I didn't see that. Thanks. $\endgroup$ Aug 4 '11 at 15:05
  • $\begingroup$ Actually you cannot apply L'H here, because the limit is the definition of the derivative of $\sin( \sin (x))$ at $x=0$. $\endgroup$
    – N. S.
    Aug 27 '12 at 16:56
  • $\begingroup$ @N.S. Why does that preclude use of L'H? $\endgroup$ Sep 9 '12 at 21:22
  • $\begingroup$ Because you USE the derivative of $\sin(\sin(x))$ to calculate ITSELF. That is circular logic.... $\endgroup$
    – N. S.
    Sep 10 '12 at 0:10
8
$\begingroup$

Note that :

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

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

$\endgroup$
2
  • 2
    $\begingroup$ Thanks Chandru, but I can not use the series expansion when I'm on chapter limits. But thanks for the extraordinary speed in responding. $\endgroup$ Aug 4 '11 at 14:57
  • $\begingroup$ @mathsalomon: You didn't mention that before :) $\endgroup$
    – user9413
    Aug 4 '11 at 14:58
7
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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