Solution
= Solution
The <Maclaurin series> give $1-\cos(x^2)=x^4/2+O(x^8)$ and $\tan x=x+O(x^3)$. Therefore $x^2\tan^2x=x^4+O(x^6)$, and cancellation of the common leading power gives
$$
\boxed{\lim_{x\to0}\frac{1-\cos(x^2)}{x^2\tan^2x}=\frac12}.
$$
The argument applies from either side of zero: both leading terms involve the even power $x^4$. The expression need not be defined at zero itself for its <limit> to exist.