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. 

Using taylor series you have
$$\color{#05f}{\frac{\sin x}{x}} = \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{2n}}{(2n+1)!} = 1 - \frac{x^2}{6} + \frac{x^4}{120} + O(x^6)$$
and $$\ln \Bigg(\color{#05f}{\frac{\sin x}{x}}\Bigg) = - \frac{x^2}{6} - O(x^4)$$
$$\lim _{x\to 0 } \frac{- \frac{x^2}{6} - O(x^4)}{x^2}= \lim_{x\to 0} -\frac{1}{6} - O(x^2) = \color{#f05}{-\frac{1}{6}}$$
Edit:
By L'hospital you have 
$$\begin{align}\require{cancel}\frac{d}{dx} \ln \Bigg(\frac{\sin x}{x}\Bigg) &= \frac{\color{#f05}{\cancel x}}{\sin x}\frac{x\cos x - \sin x}{x^{\color{#f05}{\cancel{2}}}} = \cot x - \frac{1}{x} \\&\implies \frac{d^2}{dx^2} \ln \Bigg(\frac{\sin x}{x}\Bigg) = \frac{1}{x^2} - \csc^2 x \color{#085}{\to -\frac{1}{3}} \end{align}$$
as $x \to 0$. 
Then 
$$\lim _{x\to 0 } \frac{\frac{1}{x^2} - \csc^2 x}{2}= {-\frac{1}{2}\frac{1}{3}} = \color{#f05}{-\frac{1}{6}}$$
