Solution
= Solution
The <Maclaurin series> of the <exponential function> and <sine> give
$$
e^{x^2}=1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+O(x^8),\qquad
x\sin x=x^2-\frac{x^4}{6}+\frac{x^6}{120}+O(x^8).
$$
Both the numerator and denominator first become nonzero at order $x^4$. Dividing by that common power gives the <limit>
$$
\boxed{\lim_{x\to0}\frac{1+x^2-e^{x^2}}{x\sin x-x^2}
=\lim_{x\to0}\frac{-1/2+O(x^2)}{-1/6+O(x^2)}=3.}
$$