Apply L’Hopital's rule twice. Then, we get
lim x → 0 1 − cos 2 x x 2 = lim x → 0 2 cos x sin x 2 x = lim x → 0 2 ( − sin x sin x + cos x cos x ) 2 = 1. {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {1-\cos ^{2}x}{x^{2}}}=\lim _{x\to 0}{\frac {2\cos x\sin x}{2x}}=\lim _{x\to 0}{\frac {2(-\sin x\sin x+\cos x\cos x)}{2}}=\color {blue}1.\end{aligned}}}
