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?

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 $\dfrac16$.
