## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 2} \frac{1-2^{4-x^{2}}}{2\left(\sqrt{2 x}-\sqrt{3 x^{2}-5 x+2}\right)}
$$
## Solution

Substitution:

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

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

We get:

$$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{1-2^{4-x^{2}}}{2\left(\sqrt{2 x}-\sqrt{3 x^{2}-5 x+2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2^{4-(y+2)^{2}}}{2\left(\sqrt{2(y+2)}-\sqrt{3(y+2)^{2}-5(y+2)+2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2^{4-y^{2}-4 y-4}}{2\left(\sqrt{2(y+2)}-\sqrt{3\left(y^{2}+4 y+4\right)-5 y-10+2}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{1-2^{-y(y+4)}}{2\left(\sqrt{2(y+2)}-\sqrt{3 y^{2}+12 y+12-5 y-8}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-\left(\left(e^{\ln 2}\right)^{-y(y+4)}-1\right)}{2\left(\sqrt{2(y+2)}-\sqrt{3 y^{2}+7 y+4}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-\left(e^{-y(y+4) \ln 2}-1\right)}{2\left(\sqrt{2(y+2)}-\sqrt{3 y^{2}+7 y+4}\right)}=
\end{aligned}
$$

$e^{-y(y+4) \ln 2}-1 \sim -y(y+4) \ln 2 \text{, as } y \rightarrow 0(-y(y+4) \ln 2 \rightarrow 0)$

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-(-y(y+4) \ln 2)}{2\left(\sqrt{2(y+2)}-\sqrt{3 y^{2}+7 y+4}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-(-y(y+4) \ln 2)\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}{2\left(\sqrt{2(y+2)}-\sqrt{3 y^{2}+7 y+4}\right)\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-(-y(y+4) \ln 2)\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}{2\left(2(y+2)-\left(3 y^{2}+7 y+4\right)\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y(y+4) \ln 2\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}{2\left(2 y+4-3 y^{2}-7 y-4\right)}= \\
& =\lim _{y \rightarrow 0} \frac{y(y+4) \ln 2\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}{2\left(-3 y^{2}-5 y\right)}= \\
& =\lim _{y \rightarrow 0} \frac{(y+4) \ln 2\left(\sqrt{2(y+2)}+\sqrt{3 y^{2}+7 y+4}\right)}{2(-3 y-5)}= \\
& =\frac{(0+4) \ln 2\left(\sqrt{2(0+2)}+\sqrt{3 \cdot 0^{2}+7 \cdot 0+4}\right)}{2(-3 \cdot 0-5)}= \\
& =\frac{4 \ln 2(\sqrt{4}+\sqrt{4})}{-10}=\frac{16 \ln 2}{-10}=-\frac{8 \ln 2}{5}
\end{aligned}
$$

## Problem Kuznetsov Limits 13-21