How do you calculate this limit $$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}?$$ without derivatives please. Thanks.
How do you calculate this limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\sin x)}}{x}$?
-
23Hint: ${\displaystyle {\sin(\sin(x)) \over x} = {\sin(\sin(x)) \over \sin(x)} {\sin(x) \over x}}$. – 2011-08-04
-
3Or, 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 $0$. – 2011-08-04
-
0Thanks Zarrax, is just the trick I needed. : D – 2011-08-04
-
0@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
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 $1 \cdot 1 = 1$.
-
0Yess!! thank you very much :D – 2011-08-04
-
3@J.J Zarrax might have a bone to pick with you. – 2012-02-22
-
0It 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
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 $$\lim_{x\to 0} \frac{\sin (\sin x)}{x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sin (\sin x)}{\frac{d}{dx}x} = \lim_{x \to 0} \frac{\cos(\sin x) \cos x}{1} = 1.$$
-
0Taking derivatives are not allowed :( – 2011-08-04
-
0@Chandru Oops. I didn't see that. Thanks. – 2011-08-04
-
0Actually you cannot apply L'H here, because the limit is the definition of the derivative of $\sin( \sin (x))$ at $x=0$. – 2012-08-27
-
0@N.S. Why does that preclude use of L'H? – 2012-09-09
-
0Because you USE the derivative of $\sin(\sin(x))$ to calculate ITSELF. That is circular logic.... – 2012-09-10
-
0@N.S. Why can't I just use the chain rule to conclude $\sin(\sin(x))$ is differentiable at $0$ and then apply L'H? – 2012-09-11
-
0So, you use chain rule, find that the derivative of $\sin(\sin(x))$ at 0, and then you apply the L'H to calculate again the derivative of $\sin(\sin(x))$ at 0.... – 2012-09-11
-
0@N.S. It might be redundant, but I don't think it's circular. Moreover, if one didn't notice that the given limit was the definition of $(\sin(\sin x))'$ at $x=0$, then it would be a practical way to proceed. – 2012-09-11
-
0"L'Hôpital's rule makes the problem much easier": well, I don't think it is any simpler than J.J.'s solution. – 2014-10-22
Note that :
$$\sin(\sin{x}) = \sin{x} - \frac{(\sin{x})^{3}}{3!} + \frac{(\sin{x})^{5}}{5!} + \cdots $$
$\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1$.
-
2Thanks Chandru, but I can not use the series expansion when I'm on chapter limits. But thanks for the extraordinary speed in responding. – 2011-08-04
-
0@mathsalomon: You didn't mention that before :) – 2011-08-04
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 $$ \lim_{x\to 0}\frac{\sin(\sin(x))}{x}=\lim_{x\to 0}\frac{\sin(\sin(x))}{\sin(x)}\lim_{x\to 0}\frac{\sin(x)}{x}=1 $$