Condition of the problem

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{\operatorname{tg} x-\sin x}{x(1-\cos 2 x)}$
## Solution

We will use the substitution of equivalent infinitesimals:

\(\sin x \sim x\), as \(x \rightarrow 0\)

\(1 - \cos x \sim \frac{x^2}{2}\), as \(x \rightarrow 0\)

\(\tan x \sim x\), as \(x \rightarrow 0\)

We get:

\[
\begin{aligned}
& \lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x(1 - \cos 2x)} = \left\{\frac{0}{0}\right\} = \lim_{x \rightarrow 0} \frac{\tan x - \frac{\sin x \cdot \cos x}{\cos x}}{x(1 - \cos^2 x + \sin^2 x)} = \\
& = \lim_{x \rightarrow 0} \frac{\tan x - \tan x \cdot \cos x}{x(2 \sin^2 x)} = \lim_{x \rightarrow 0} \frac{\tan x (1 - \cos x)}{x(2 x^2)} = \\
& = \lim_{x \rightarrow 0} \frac{x \cdot \frac{x^2}{2}}{x(2 x^2)} = \lim_{x \rightarrow 0} \frac{\frac{1}{2}}{2} = \frac{1}{4}
\end{aligned}
\]

## Problem Kuznetsov Limits 12-27