146. Find $\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}$.
Solution. Method I. Here we have an indeterminate form of the type $0 / 0$. Applying a known trigonometric formula and performing elementary transformations, we get

$$
\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{2}}=\lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{\sin (x / 2)}{x / 2} \cdot \frac{\sin (x / 2)}{x / 2}=\frac{1}{2}
$$

Method II. Transform the numerator as follows:

$$
\frac{1-\cos x}{x^{2}}=\frac{(1-\cos x)(1+\cos x)}{x^{2}(1+\cos x)}=\frac{1-\cos ^{2} x}{x^{2}(1+\cos x)}=\left(\frac{\sin x}{x}\right)^{2} \cdot \frac{1}{1+\cos x}
$$

Therefore,

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{2} \cdot \lim _{x \rightarrow 0} \frac{1}{1+\cos x}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}\right)^{2} \cdot \frac{1}{1+\lim _{x \rightarrow 0} \cos x}=1^{2} \cdot \frac{1}{1+1}=\frac{1}{2}
\end{aligned}
$$