Solution
= Solution
The <Taylor expansions> at zero give $1-e^x=-x-x^2/2+O(x^3)$ and $1-\cos x=x^2/2+O(x^4)$. Squaring the first expression gives $x^2+O(x^3)$, so canceling the leading quadratic power yields
$$
\boxed{\lim_{x\to0}\frac{(1-e^x)^2}{1-\cos x}=2.}
$$
Both numerator and denominator vanish, but their leading nonzero coefficients have ratio two.