## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow \frac{1}{2}} \frac{\ln (4 x-1)}{\sqrt{1-\cos \pi x}-1}$
## Solution

Substitution:

$x=y+\frac{1}{2} \Rightarrow y=x-\frac{1}{2}$

$x \rightarrow \frac{1}{2} \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow \frac{1}{2}} \frac{\ln (4 x-1)}{\sqrt{1-\cos \pi x}-1}=\lim _{y \rightarrow 0} \frac{\ln \left(4\left(y+\frac{1}{2}\right)-1\right)}{\sqrt{1-\cos \left(\pi\left(y+\frac{1}{2}\right)\right)}-1}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (4 y+2-1)}{\sqrt{1-\cos \left(\pi y+\frac{\pi}{2}\right)}-1}=\lim _{y \rightarrow 0} \frac{\ln (4 y+1)}{\sqrt{1+\sin \pi y}-1}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln (4 y+1) \sim 4 y_{\text {, as }} y \rightarrow 0(4 y \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{4 y}{\sqrt{1+\sin \pi y}-1}= \\
& =\lim _{y \rightarrow 0} \frac{4 y(\sqrt{1+\sin \pi y}+1)}{(\sqrt{1+\sin \pi y}-1)(\sqrt{1+\sin \pi y}+1)}= \\
& =\lim _{y \rightarrow 0} \frac{4 y(\sqrt{1+\sin \pi y}+1)}{1+\sin \pi y-1}= \\
& =\lim _{y \rightarrow 0} \frac{4 y(\sqrt{1+\sin \pi y}+1)}{\sin \pi y}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\sin \pi y \sim \pi y{ }_{, \text {as }} y \rightarrow 0(\pi y \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{4 y(\sqrt{1+\sin \pi y}+1)}{\pi y}=\lim _{y \rightarrow 0} \frac{4(\sqrt{1+\sin \pi y}+1)}{\pi}= \\
& =\frac{4(\sqrt{1+\sin \pi 0}+1)}{\pi}=\frac{4(\sqrt{1+0}+1)}{\pi}= \\
& =\frac{4(\sqrt{1+0}+1)}{\pi}=\frac{8}{\pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-9