# Find $\lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$. Is my approach correct?

Find: $$L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$$

My approach:

Because of the fact that the above limit is evaluated as $\frac{0}{0}$, we might want to try the De L' Hospital rule, but that would lead to a more complex limit which is also of the form $\frac{0}{0}$.

What I tried is: $$L = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right)$$ Then, if the limits $$L_1 = \lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{1-\frac{\sin(x)}{x}},$$

$$L_2 = \lim_{x\to0}\frac{1}{x^2}\left(1-\frac{\sin(x)}{x}\right)$$ exist, then $L=L_1L_2$.

For the first one, by making the substitution $u=1-\frac{\sin(x)}{x}$, we have $$L_1 = \lim_{u\to u_0}\frac{\sin(u)}{u},$$ where $$u_0 = \lim_{x\to0}\left(1-\frac{\sin(x)}{x}\right)=0.$$ Consequently, $$L_1 = \lim_{u\to0}\frac{\sin(u)}{u}=1.$$

Moreover, for the second limit, we apply the De L' Hospital rule twice and we find $L_2=\frac{1}{6}$.

Finally, $L=1\frac{1}{6}=\frac{1}{6}$.

Is this correct?

• The answer is right. The $1/6$ is clear from the power series expansion of $\frac{\sin x}{x}$. – André Nicolas Dec 7 '15 at 20:51
• Thank you very much @AndréNicolas! – nullgeppetto Dec 7 '15 at 20:52
• You are welcome. That was a nice trick, multiplying and dividing by $1-\frac{\sin x}{x}$. Using L'Hospital's Rule afterwards is OK, but in general when possible I get a better feel of what's happening from the power series than from L'Hospital's Rule. – André Nicolas Dec 7 '15 at 20:57
• Better to do $L_2$ as the limit of $\frac{x-\sin x}{x^3}$, which requires L'Hopital three times. You might have to apply some additional knowledge your way. – Thomas Andrews Dec 7 '15 at 20:57
• Cleverly done ! – Yves Daoust Dec 8 '15 at 11:14

In a slightly different way, using the Taylor expansion, as $x \to 0$, $$\sin x=x-\frac{x^3}6+O(x^5)$$ gives $$1-\frac{\sin x}x=\frac{x^2}6+O(x^4)$$ then $$\sin \left( 1-\frac{\sin x}x\right)=\frac{x^2}6+O(x^4)$$ and

$$\frac{\sin \left( 1-\frac{\sin x}x\right)}{x^2}=\frac16+O(x^2)$$

from which one may conclude easily.

By L' Hospital anyway:

$$\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$$ yields

$$\cos\left(1-\frac{\sin(x)}x\right)\frac{\sin(x)-x\cos(x)}{2x^3}.$$

The first factor has limit $1$ and can be ignored.

Then with L'Hospital again:

$$\frac{x\sin(x)}{6x^2},$$

which clearly tends to $\dfrac16$.

• Wasn't so horrible after all. – Yves Daoust Dec 8 '15 at 11:29