## Problem Statement

Calculate the limit of the function:

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

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

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

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

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

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

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

Using the substitution of equivalent infinitesimals:

$\ln \left(1-x \cdot \frac{3^{x}-2^{x}}{1+x \cdot 3^{x}}\right) \sim -x \cdot \frac{3^{x}-2^{x}}{1+x \cdot 3^{x}}$, as $\quad x \rightarrow 0\left(-x \cdot \frac{3^{x}-2^{x}}{1+x \cdot 3^{x}} \rightarrow 0\right)$

We get:

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

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

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

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

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

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

Using the substitution of equivalent infinitesimals:

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

$e^{x \cdot \ln 2}-1 \sim x \cdot \ln 2$, as $x \rightarrow 0(x \cdot \ln 2 \rightarrow 0)$

We get:

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

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

$=\exp \left\{-\frac{\ln 3}{1+0 \cdot 3^{0}}+\frac{\ln 2}{1+0 \cdot 3^{0}}\right\}=$

$$
=\exp \{-\ln 3+\ln 2\}=\exp \left\{\ln \frac{2}{3}\right\}=e^{\ln \frac{2}{3}}=\frac{2}{3}
$$

## Problem Kuznetsov Limits 17-3