limits - Find $lim_{xto0}frac{sinleft(1-frac{sin(x)}{x}right)}{x^2}$. 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?

Answer

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.