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. 

Hint: Use finite expansion for $sin x$ in the neighberhood of zero! Will lead you to $ln(1-x^2/6)$ ! Using finite expansion again on $ln(-x^2/6+1)$ will lead to $-x^2/6$! You can evaluate the limit then which will then lead to $-1/6$ i guess! Sorry for my bad language.
