Solution
= Solution
The cancellation of the quadratic terms makes the fourth-order terms decisive. The <Taylor expansions> give
$$
1+x^2-e^{x^2}=-\frac{x^4}{2}-\frac{x^6}{6}+O(x^8),\qquad x\sin x-x^2=-\frac{x^4}{6}+\frac{x^6}{120}+O(x^8).
$$
Canceling $x^4$ for $x\ne0$ and taking the <limit> yields
$$
\boxed{\lim_{x\to0}\frac{1+x^2-e^{x^2}}{x\sin x-x^2}=3.}
$$