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. 
 A: We want$$
L = \lim_{x\to 0} \frac{\ln(\frac{\sin x}{x})}{x^2}
$$
Since the top approaches $\ln(1) = 0$ and the bottom also approaches $0$, we may use L'Hopital:
$$
L = \lim_{x\to 0}{\frac{(\frac{x}{\sin x})(\frac{x \cos x - \sin x}{x^2})}{2x}} = \lim_{x\to 0}\frac{x \cos x - \sin x}{2x^2\sin x}
$$
Again the top and  bottom both approach $0$, so again we may use L'Hopital.
$$
L = \lim_{x \to 0} \frac{\cos x - x \sin x - \cos x}{4x \sin x + 2x^2 \cos x} = \lim_{x \to 0} \frac{-\sin x}{4 \sin x + 2x \cos x}
$$
Again the top and bottom both approach $0$, so we may use L'Hopital for a third time:
$$
L = \lim_{x \to 0} \frac{-\cos x}{4 \cos x + 2\cos x - 2x\sin x} = -\frac{1}{6}.
$$
A: 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 $\,-\dfrac16$.
A: 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.
A: 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 
$$\lim _{x\to 0 } \frac{\frac{1}{x^2} - \csc^2 x}{2}= {-\frac{1}{2}\frac{1}{3}} = \color{#f05}{-\frac{1}{6}}$$
A: It is possible to compute
\begin{equation*}
\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{\ln \left( \frac{\sin x%
}{x}\right) }{x^{2}}
\end{equation*}
only by making use of some basic limits: 
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6} \\
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1.
\end{eqnarray*}
No l'Hospital's rule nor Taylor series are required. Let
\begin{eqnarray*}
f(x) &=&\frac{1}{x^{2}}\ln \frac{\sin x}{x} \\
&=&\frac{1}{x^{2}}\ln \left( 1+[\frac{\sin x}{x}-1]\right)  \\
&=&\frac{\left[ \frac{\sin x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+%
\left[ \frac{\sin x}{x}-1\right] \right) }{\left[ \frac{\sin x}{x}-1\right] }
\\
&=&\frac{\sin x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{%
\sin x}{x}-1
\end{eqnarray*}
since 
\begin{equation*}
\lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\sin x}{x}-1)=1-1=0
\end{equation*}
then 
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{%
\ln (1+u)}{u}=1,
\end{equation*}
and therefore 
\begin{equation*}
\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\cdot
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=-\frac{1}{6}\cdot 1=-\frac{1}{%
6}.
\end{equation*}
A: In these cases using a Taylor Series approach usually gets you to the solution: The denominator will be expanded as:
$$\frac{-x^2}{6} + O(x^3)$$
 Thus dividing by $x^2$ as $x\rightarrow 0$ leaves you with $\frac{-1}{6}$.
A: Recall l'Hospital's rule:  If the limit of $f(x)/g(x)$ for $x \rightarrow 0$ is undefined, and if the limit of $f'(x)/g'(x)$ is also undefined, then calculate $f''(x)/g''(x)$.  In this case:
$\lim_{x\to 0} \, \frac{\partial ^2}{\partial x^2}\log \left(\frac{\sin (x)}{x}\right) = -1/3$
and
$\lim_{x\to 0} \, \frac{\partial ^2x^2}{\partial x^2} = 2$,
so the limit of the full term is $-1/6$.
