## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)^{\frac{1}{x^2}}
$$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{1+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)^{\frac{1}{\tan^{2} x}}=$

$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\left(1+x \cdot 3^{x}\right) /\left(1+x \cdot 7^{x}\right)\right)}\right)^{\frac{1}{\tan^{2} x}}=$

$=\lim _{x \rightarrow 0} e^{\frac{1}{\tan^{2} x} \ln \left(\left(1+x \cdot 3^{x}\right) /\left(1+x \cdot 7^{x}\right)\right)}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\tan^{2} x} \ln \left(\left(1+x \cdot 3^{x}\right) /\left(1+x \cdot 7^{x}\right)\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\tan^{2} x} \ln \left(\frac{1+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\tan^{2} x} \ln \left(\frac{1+x \cdot 7^{x}-x \cdot 7^{x}+x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\tan^{2} x} \ln \left(1-\frac{x \cdot 7^{x}-x \cdot 3^{x}}{1+x \cdot 7^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\tan^{2} x} \ln \left(1-x \cdot \frac{7^{x}-3^{x}}{1+x \cdot 7^{x}}\right)\right\}=$

Using the substitution of equivalent infinitesimals:

$\ln \left(1-x \cdot \frac{7^{x}-3^{x}}{1+x \cdot 7^{x}}\right) \sim -x \cdot \frac{7^{x}-3^{x}}{1+x \cdot 7^{x}}$, as $x \rightarrow 0\left(-x \cdot \frac{7^{x}-3^{x}}{1+x \cdot 7^{x}} \rightarrow 0\right)$ $\tan x \sim x$, as $x \rightarrow 0$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{x^{2}}\left(-x \cdot \frac{7^{x}-3^{x}}{1+x \cdot 7^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{7^{x}-3^{x}}{x \cdot\left(1+x \cdot 7^{x}\right)}\right\}=$

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

Using the substitution of equivalent infinitesimals:

$e^{x \cdot \ln 7}-1 \sim x \cdot \ln 7$, as $x \rightarrow 0(x \cdot \ln 7 \rightarrow 0)$

$e^{x \cdot \ln 3}-1 \sim x \cdot \ln 3$, as $x \rightarrow 0(x \cdot \ln 3 \rightarrow 0)$

We get:

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

## Problem Kuznetsov Limits 17-27