## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{4 x^{2}}
$$
## Solution

We will use the substitution of equivalent infinitesimals:

$$
1-\cos x \sim \frac{x^{2}}{2}, \text { as } x \rightarrow 0
$$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\cos ^{3} x}{4 x^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{(1-\cos x)\left(1+\cos x+\cos ^{2} x\right)}{4 x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{x^{2}}{2}\left(1+\cos x+\cos ^{2} x\right)}{4 x^{2}}=\lim _{x \rightarrow 0} \frac{1+\cos x+\cos ^{2} x}{8}= \\
& =\frac{1+1+1^{2}}{8}=\frac{3}{8}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-7