## Problem Statement

Calculate the limit of the function:
$\lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}$
## Solution

Substitution:

$x=y+3 \Rightarrow y=x-3$

$x \rightarrow 3 \Rightarrow y \rightarrow 0$

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{\sin \left(\sqrt{2 x^{2}-3 x-5}-\sqrt{1+x}\right)}{\ln (x-1)-\ln (x+1)+\ln 2}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \left(\sqrt{2(y+3)^{2}-3(y+3)-5}-\sqrt{1+(y+3))}\right.}{\ln ((y+3)-1)-\ln ((y+3)+1)+\ln 2}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \left(\sqrt{2 y^{2}+12 y+18-3 y-9-5}-\sqrt{y+4}\right)}{\ln (y+2)-\ln (y+4)+\ln 2}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \left(\sqrt{2 y^{2}+9 y+4}-\sqrt{y+4}\right)}{\ln \frac{2(y+2)}{y+4}}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \frac{\left(\sqrt{2 y^{2}+9 y+4}-\sqrt{y+4}\right)\left(\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}\right)}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}}}{\ln \left(1+\frac{y}{y+4}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \frac{2 y^{2}+9 y+4-(y+4)}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}}}{\ln \left(1+\frac{y}{y+4}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{\sin \frac{2 y^{2}+8 y}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}}}{\ln \left(1+\frac{y}{y+4}\right)}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\begin{aligned}
& \sin \frac{2 y^{2}+8 y}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}} \sim \frac{2 y^{2}+8 y}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}} \\
& \text {, as } \\
& y \rightarrow 0\left(\frac{2 y^{2}+8 y}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}} \rightarrow 0\right) \\
& \ln \left(1+\frac{y}{y+4}\right) \sim \frac{y}{y+4}{ }_{\text {, as }} y \rightarrow 0\left(\frac{y}{y+4} \rightarrow 0\right)
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{\frac{2 y^{2}+8 y}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}}}{\frac{y}{y+4}}= \\
& =\lim _{y \rightarrow 0} \frac{(2 y+8)(y+4)}{\sqrt{2 y^{2}+9 y+4}+\sqrt{y+4}}= \\
& =\frac{(2 \cdot 0+8)(0+4)}{\sqrt{2 \cdot 0^{2}+9 \cdot 0+4}+\sqrt{0+4}}=\frac{8 \cdot 4}{\sqrt{4}+\sqrt{4}}=\frac{8 \cdot 4}{2+2}=8
\end{aligned}
$$

## Problem Kuznetsov Limits 14-7