Condition of the problem

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 7 x-\cos 3 x}$
## Solution

We will use the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& 1-\cos 2 x \sim \frac{(2 x)^{2}}{2}, \text { as } x \rightarrow 0(2 x \rightarrow 0) \\
& \sin 5 x \sim 5 x, \text { as } x \rightarrow 0(5 x \rightarrow 0) \\
& \sin 2 x \sim 2 x, \text { as } x \rightarrow 0(2 x \rightarrow 0)
\end{aligned}
$$

We obtain:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 1 x-\cos 3 x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \frac{\frac{4 x^{2}}{2}}{-2 \sin \frac{7 x+3 x}{2} \sin \frac{7 x-3 x}{2}}= \\
& =\lim _{x \rightarrow 0} \frac{2 x^{2}}{-2 \sin 5 x \cdot \sin 2 x}=\lim _{x \rightarrow 0} \frac{2 x^{2}}{-2 \cdot 5 x \cdot 2 x}= \\
& =\lim _{x \rightarrow 0} \frac{1}{-5 \cdot 2}=-\frac{1}{10}
\end{aligned}
$$

## Problem Kuznetsov Limits 12-4