## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 1} \frac{\sqrt{2^{x}+7}-\sqrt{2^{x+1}+5}}{x^{3}-1}
$$
## Solution

Substitution:

$$
\begin{aligned}
& x=y+1 \Rightarrow y=x-1 \\
& x \rightarrow 1 \Rightarrow y \rightarrow 0
\end{aligned}
$$

We get:

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

Using the substitution of equivalent infinitesimals:

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

We get:

$$
\begin{aligned}
& =\lim _{y \rightarrow 0} \frac{-2 y \ln 2}{y\left(y^{2}+3 y+3\right)\left(\sqrt{2^{y+1}+7}+\sqrt{2^{y+2}+5}\right)}= \\
& =\lim _{y \rightarrow 0} \frac{-2 \ln 2}{\left(y^{2}+3 y+3\right)\left(\sqrt{2^{y+1}+7}+\sqrt{2^{y+2}+5}\right)}= \\
& =\frac{-2 \ln 2}{\left(0^{2}+3 \cdot 0+3\right)\left(\sqrt{2^{0+1}+7}+\sqrt{2^{0+2}+5}\right)}= \\
& =\frac{-2 \ln 2}{3\left(\sqrt{2+7}+\sqrt{2^{2}+5}\right)}=\frac{-2 \ln 2}{3(3+3)}=\frac{-2 \ln 2}{3 \cdot 6}=
\end{aligned}
$$

$=-\frac{\ln 2}{9}$

## Problem Kuznetsov Limits 14-21