 Solution

The Maclaurin series give $1-\cos (x^2)=x^4/2+O(x^8)$ and $\tan x=x+O(x^3)$. Therefore $x^2{\tan }^{2}x=x^4+O(x^6)$, and cancellation of the common leading power givesThe 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.