## Solution

Substitution:

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

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

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{\ln (x-\sqrt[3]{2 x-3})}{\sin \left(\frac{\pi x}{2}\right)-\sin ((x-1) \pi)}=\lim _{y \rightarrow 0} \frac{\ln ((y+2)-\sqrt[3]{2(y+2)-3})}{\sin \left(\frac{\pi(y+2)}{2}\right)-\sin (((y+2)-1) \pi)}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (y+2-\sqrt[3]{2 y+1})}{\sin \left(\frac{\pi y}{2}+\pi\right)-\sin (y \pi+\pi)}=\lim _{y \rightarrow 0} \frac{\ln (1+(y+1-\sqrt[3]{2 y+1}))}{\sin y \pi-\sin \frac{\pi y}{2}}= \\
& =\lim _{y \rightarrow 0} \frac{\ln (1+(y+1-\sqrt[3]{2 y+1}))}{2 \sin \frac{y \pi-\frac{\pi y}{2}}{2} \cos \frac{y \pi+\frac{\pi y}{2}}{2}}=\lim _{y \rightarrow 0} \frac{\ln (1+(y+1-\sqrt[3]{2 y+1}))}{2 \sin \frac{\pi y}{4} \cos \frac{3 \pi y}{4}}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$\ln (1+(y+1-\sqrt[3]{2 y+1})) \sim y+1-\sqrt[3]{2 y+1}$, as $y \rightarrow 0((y+1-\sqrt[3]{2 y+1}) \rightarrow 0)$

$\sin \frac{\pi y}{4} \sim \frac{\pi y}{4}$, as $y \rightarrow 0\left(\frac{\pi y}{4} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{y+1-\sqrt[3]{2 y+1}}{2 \frac{\pi y}{4} \cos \frac{3 \pi y}{4}}= \\
& =\lim _{y \rightarrow 0} \frac{(y+1-\sqrt[3]{2 y+1})\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}{\frac{\pi y}{2} \cos \left(\frac{3 \pi y}{4}\right)\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{(y+1)^{3}-(2 y+1)}{\frac{\pi y}{2} \cos \left(\frac{3 \pi y}{4}\right)\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{3}+3 y^{2}+3 y+1-2 y-1}{\frac{\pi y}{2} \cos \left(\frac{3 \pi y}{4}\right)\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{3}+3 y^{2}+y}{\frac{\pi y}{2} \cos \left(3 \frac{\pi y}{4}\right)\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y^{2}+3 y+1}{\frac{\pi}{2} \cos \left(3 \frac{\pi y}{4}\right)\left((y+1)^{2}+(y+1) \sqrt[3]{2 y+1}+\sqrt[3]{(2 y+1)^{2}}\right)}= \\
& =\frac{0^{2}+3 \cdot 0+1}{\frac{\pi}{2} \cos \left(\frac{3 \pi \cdot 0}{4}\right)\left((0+1)^{2}+(0+1) \sqrt[3]{2 \cdot 0+1}+\sqrt[3]{(2 \cdot 0+1)^{2}}\right)}= \\
& =\frac{1}{\frac{\pi}{2} \cdot 1 \cdot\left(1+1 \cdot \sqrt[3]{1}+\sqrt[3]{1^{2}}\right)}=\frac{1}{\frac{3 \pi}{2}}=\frac{2}{3 \pi}
\end{aligned}
$$

## Problem Kuznetsov Limits 14-3
## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 2} \frac{\ln (x-\sqrt[3]{2 x-3})}{\sin \left(\frac{\pi x}{2}\right)-\sin ((x-1) \pi)}
$$