## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow a} \frac{a^{\left(x^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{x}{a}\right)}$
## Solution

Substitution:

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

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

We get:

$\lim _{x \rightarrow a} \frac{a^{\left(x^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{x}{a}\right)}=\lim _{y \rightarrow 0} \frac{a^{\left((y+a)^{2}-a^{2}\right)}-1}{\tan \ln \left(\frac{y+a}{a}\right)}=$

$=\lim _{y \rightarrow 0} \frac{a^{\left(y^{2}+2 a y+a^{2}-a^{2}\right)}-1}{\tan \ln \left(1+\frac{y}{a}\right)}=$

$=\lim _{y \rightarrow 0} \frac{a^{\left(y^{2}+2 a y\right)}-1}{\tan \ln \left(1+\frac{y}{a}\right)}=\lim _{y \rightarrow 0} \frac{\left(e^{\ln a}\right)^{\left(y^{2}+2 a y\right)}-1}{\tan \ln \left(1+\frac{y}{a}\right)}=$

$=\lim _{y \rightarrow 0} \frac{e^{\ln a \cdot\left(y^{2}+2 a y\right)}-1}{\tan \ln \left(1+\frac{y}{a}\right)}=$

Using the substitution of equivalent infinitesimals:
$e^{\ln a \cdot\left(y^{2}+2 a y\right)}-1 \sim \ln a \cdot\left(y^{2}+2 a y\right)_{\text {, as }} y \rightarrow 0\left(\ln a \cdot\left(y^{2}+2 a y\right) \rightarrow 0\right)$

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

We get:

$=\lim _{y \rightarrow 0} \frac{\ln a \cdot\left(y^{2}+2 a y\right)}{\tan \frac{y}{a}}=$

Using the substitution of equivalent infinitesimals:

$\tan \frac{y}{a} \sim \frac{y}{a}_{\text {, as }} y \rightarrow 0\left(\frac{y}{a} \rightarrow 0\right)$

We get:

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

## Problem Kuznetsov Limits 14-27