Calculating the limit: $\lim \limits_{x \to 0}$ $\frac{\ln(\frac{\sin x}{x})}{x^2}. $ How do I calculate $$\lim \limits_{x \to 0} \dfrac{\ln\left(\dfrac{\sin x}{x}\right)}{x^2}\text{?}$$ 
I thought about using L'Hôpital's rule, applying on "$\frac00$," but then I thought about $\frac{\sin x}{x}$  which is inside the $\ln$: it's not constant as $x$ goes to $0$, then I thought that maybe this what caused that my calculating of the limit wasn't true. 
Can someone clarify how we calculate the limit here? 
Note: I'd like to see a solution using L'Hôpital's rule. 

As $\dfrac{\sin x}{x}=1-\dfrac{x^2}6+o(x^3) $, we have:
$$\frac{\ln\Bigl(\cfrac{\sin x}{x}\Bigr)}{x^2}=\frac{\ln\Bigl(1-\cfrac{x^2}6+o(x^3)\Bigr)}{x^2}=\frac{-\dfrac{x^2}6+o(x^3)}{x^2}=-\frac16+o(x),$$
which proves the limit is $\,-\dfrac16$.
