## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 2} \frac{\tan x - \tan 2}{\sin (\ln (x-1))}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\tan x - \tan 2}{\sin (\ln (x-1))} = \lim _{x \rightarrow 2} \frac{\frac{\sin (x-2)}{\cos x \cdot \cos 2}}{\sin (\ln (x-1))} = \\
& = \lim _{x \rightarrow 2} \frac{\sin (x-2)}{\cos x \cdot \cos 2 \cdot \sin (\ln (x-1))} =
\end{aligned}
$$

Substitution:

$x = y + 2 \Rightarrow y = x - 2$

$x \rightarrow 2 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& = \lim _{y \rightarrow 0} \frac{\sin (y + 2 - 2)}{\cos (y + 2) \cdot \cos 2 \cdot \sin (\ln (y + 2 - 1))} = \\
& = \lim _{y \rightarrow 0} \frac{\sin y}{\cos (y + 2) \cdot \cos 2 \cdot \sin (\ln (1 + y))} =
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:
$\ln (1 + y) \sim y$, as $y \rightarrow 0$

We get:

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

## Problem Kuznetsov Limits 14-4