Finding $\lim_{x\to 0}\frac{\sin(x+x^3/6)-x}{x^5}$ I'm trying to find the limit of this expression:

$$\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}$$

My solution is as follows:
$$
\begin{align}
\lim_{x\to0}\frac{\sin\left(x+x^3/6\right)-x}{x^5}&=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[\frac{(x+x^3/6)\sin(x+x^3/6)}{(x+x^3/6)}-x\right]\\
&=\lim_{x\to0}\frac{1}{x^5}\cdot\!\!\left[(x+x^3/6)-x\right]\\
&=\lim_{x\to0}\frac{x^3}{6x^5}\\
&=\lim_{x\to0}\frac{1}{6x^2}=+\infty
\end{align}
$$
But Wolfram Alpha finds the limit as $-\frac{3}{40}$. Where is my mistake? 

For the error, see the comments and other answers which address it already. If you are familiar with Taylor series, you can compute your limit as follows, observing that 
$$
\sin y \operatorname*{=}_{y\to 0} y - \frac{y^3}{6} + \frac{y^5}{120} +o(y^5)\ .
$$
$$\begin{align}
\sin (x+\frac{x^3}{6}) &\operatorname*{=}_{x\to 0} x+\frac{x^3}{6} - \frac{(x+\frac{x^3}{6})^3}{6} + \frac{(x+\frac{x^3}{6}^5)}{120} +o(x^5) \\
&= x+\frac{x^3}{6} -\left(\frac{x^3}{6}+\frac{x^5}{12}\right) + \frac{x^5}{120 }  +o(x^5) \\
&= x-\frac{3}{40}x^5 +o(x^5)
\end{align}$$
where we ignore terms $x^k$ for $k > 5$ (as they get "swallowed" in the $o(x^5)$) when developing the $(\cdot)^3$ and $(\cdot)^5$. Note that we need to go at least to the second term of the sinus (in $y^3$) as the first will be canceled out by the $-x$ that we are about to add:
$$\begin{align}
\frac{\sin (x+\frac{x^3}{6}) - x}{x^5}&\operatorname*{=}_{x\to 0} \frac{x-\frac{3}{40}x^5 +o(x^5) - x}{x^5} = \frac{-\frac{3}{40}x^5 +o(x^5)}{x^5} = -\frac{3}{40} + o(1)
\end{align}$$
yielding the limit.
