## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow a} \frac{\tan x - \tan a}{\ln x - \ln a}$
## Solution

$\lim _{x \rightarrow a} \frac{\tan x-\tan a}{\ln x-\ln a}=\lim _{x \rightarrow a} \frac{\left(\frac{\sin (x-a)}{\cos x \cdot \cos a}\right)}{\ln \frac{x}{a}}=$

Substitution:

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

$x \rightarrow a \Rightarrow y \rightarrow 0$

We get:

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

Using the substitution of equivalent infinitesimals:

$\sin y \sim y \text{, as } y \rightarrow 0$

$\ln \left(1+\frac{y}{a}\right) \sim \frac{y}{a} \text{ as } y \rightarrow 0 \left(\frac{y}{a} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 16-4