Using trigonometric identity 1 − cos 2 x = sin 2 x {\displaystyle 1-\cos ^{2}x=\sin ^{2}x} , the given function can be written as
1 − cos 2 x x 2 = sin 2 x x 2 = ( sin x x ) 2 . {\displaystyle {\begin{aligned}{\frac {1-\cos ^{2}x}{x^{2}}}&={\frac {\sin ^{2}x}{x^{2}}}=\left({\frac {\sin x}{x}}\right)^{2}.\end{aligned}}}
Since we have lim x → 0 sin x x = lim x → 0 sin x − sin 0 x − 0 = d d x ( sin x ) | x = 0 = cos x | x = 0 = 1 {\displaystyle \lim _{x\to 0}{\frac {\sin x}{x}}=\lim _{x\to 0}{\frac {\sin x-\sin 0}{x-0}}=\left.{\frac {d}{dx}}(\sin x)\right|_{x=0}=\left.\cos x\right|_{x=0}=1} ,
by the continuity of the square function, we obtain the desired limit as
lim x → 0 1 − cos 2 x x 2 = lim x → 0 ( sin x x ) 2 = ( lim x → 0 sin x x ) 2 = 1. {\displaystyle {\begin{aligned}\lim _{x\to 0}{\frac {1-\cos ^{2}x}{x^{2}}}=\lim _{x\to 0}\left({\frac {\sin x}{x}}\right)^{2}=\left(\lim _{x\to 0}{\frac {\sin x}{x}}\right)^{2}=\color {blue}1.\end{aligned}}}