<s>[INST] 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? 
 [/INST] 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.
</s>