Q:

Solving a limit using MacLaurin series

I want to find
$$ \lim_{x\to0} \frac{(e^{-x^2}-1)\sin x }{x\ln(1+x^2)}$$
using a Maclaurin series and not using the l'Hôpital's rule.
However I can't seem to get it right.
Thanks for any possible answers.

A:

$e^{-x^2}-1=-x^2+o(x^3), x \to 0$
$\ln (1 + x^2)=x^2+o(x^3), x \to 0$
$\sin x = x+o(x^2),  \ x \to 0$
The limit is -1.

