## Problem Statement

Calculate the limit of the function:

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

Substitution:

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

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

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow a \pi} \frac{\ln \left(\cos \left(\frac{x}{a}\right)+2\right)}{a^{a^{2} \pi^{2} / x^{2}-a \pi / x}-a^{a \pi / x-1}}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(\cos \left(\frac{y+a \pi}{a}\right)+2\right)}{a^{a^{2} \pi^{2} /(y+a \pi)^{2}-a \pi /(y+a \pi)}-a^{a \pi /(y+a \pi)-1}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+1-\cos \left(\frac{y}{a}\right)\right)}{a^{a \pi /(y+a \pi)-1}\left(a^{a^{2} \pi^{2} /(y+a \pi)^{2}-2 a \pi /(y+a \pi)+1}-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+1-\cos \left(\frac{y}{a}\right)\right)}{a^{a \pi /(y+a \pi)-1}\left(a^{(a \pi /(y+a \pi)-1)^{2}}-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+1-\cos \left(\frac{y}{a}\right)\right)}{a^{a \pi /(y+a \pi)-1}\left(a^{(a \pi /(y+a \pi)-(y+a \pi) /(y+a \pi))^{2}}-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+1-\cos \left(\frac{y}{a}\right)\right)}{a^{a \pi /(y+a \pi)-1}\left(\left(e^{\ln a}\right)^{(-y /(y+a \pi))^{2}}-1\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln \left(1+1-\cos \left(\frac{y}{a}\right)\right)}{a^{a \pi /(y+a \pi)-1}\left(e^{\ln a \cdot y^{2} /(y+a \pi)^{2}}-1\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \ln \left(1+1-\cos \left(\frac{y}{a}\right)\right) \sim 1-\cos \left(\frac{y}{a}\right)_{, \text {as }} y \rightarrow 0\left(1-\cos \left(\frac{y}{a}\right) \rightarrow 0\right) \\
& e^{\ln a \cdot y^{2} /(y+a \pi)^{2}}-1 \sim \ln a \cdot y^{2} /(y+a \pi)^{2}, \text { as } y \rightarrow 0\left(\ln a \cdot y^{2} /(y+a \pi)^{2} \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{1-\cos \left(\frac{y}{a}\right)}{a^{a \pi /(y+a \pi)-1}\left(\frac{\ln a \cdot y^{2}}{(y+a \pi)^{2}}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\left(1-\cos \left(\frac{y}{a}\right)\right)(y+a \pi)^{2}}{a^{a \pi /(y+a \pi)-1}\left(\ln a \cdot y^{2}\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
1-\cos \left(\frac{y}{a}\right) \sim \frac{1}{2}\left(\frac{y}{a}\right)_{, \text {as }}^{2} y \rightarrow 0\left(\frac{y}{a} \rightarrow 0\right)
$$

## We get:

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

$=\lim _{y \rightarrow 0} \frac{(y+a \pi)^{2}}{2 a^{2} \cdot a^{a \pi /(y+a \pi)-1}(\ln a)}=$

$=\frac{(0+a \pi)^{2}}{2 a^{2} \cdot a^{a \pi /(0+a \pi)-1}(\ln a)}=$
$=\frac{a^{2} \cdot \pi^{2}}{2 a^{2} \cdot a^{a \pi / a \pi-1}(\ln a)}=\frac{\pi^{2}}{2 \cdot a^{0}(\ln a)}=\frac{\pi^{2}}{2 \ln a}$

## Problem Kuznetsov Limits 14-29