How to compute $\lim_{x \to 0} (\frac{x^5 e^{-1/x^2}+x/2 - \sin(x/2))}{x^3})$? I have a problem with this limit. I have no idea where is the problem.
Can you correct my mistake? Thanks
$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$
I used the developments of McLaurin $e^x$ and $\sin x$
$$\lim\limits_{x \to 0} \left(\frac{x^5 (1-\frac{1}{x^2}+\frac{1}{2x^4})+\frac{x}{2} - ((\frac{x}{2})-(\frac{x^3}{48}))}{x^3}\right) = 
\lim\limits_{x \to 0} \left(\frac{x^5-x^3+ \frac{x}{2} +\frac{x}{2} - \frac{x}{2}+\frac{x^3}{48}}{x^3}\right)=$$ 
$$\lim\limits_{x \to 0} \left(\frac{-x^3+\frac{x^3}{48}}{x^3}\right)= 
\lim\limits_{x \to 0} \left(\frac{-\frac{47x^3}{48}}{x^3}\right)=\lim\limits_{x \to 0} \left(-\frac{47x^3}{48x^3}\right)= -\frac{47}{48}$$ 
but the result is wrong.

One may write, as $x \to 0$,
$$
\begin{align}
\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}&=\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \left(\frac{x}2-\frac{x^3}{48}+O(x^5)\right)}{x^3}\\\\
&=\frac{x^5 e^{-\frac1{x^2}} + \frac{x^3}{48}+O(x^5)}{x^3}\\\\
&=\frac{x^2 e^{-\frac1{x^2}} + \frac{1}{48}+O(x^2)}{1}\\\\
&=  \frac{1}{48}+O(x^2)\\\\
\end{align}
$$ giving $\dfrac{1}{48}$ for the sought limit, where we have used $\displaystyle x^2 e^{-\frac1{x^2}}=O(x^2)$ (to say the least).
