## Problem Statement

Calculate the limit of the function:

$$
\lim _{x \rightarrow 0}\left(\frac{1+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)^{\frac{1}{\sin ^{3} x}}
$$
## Solution

$\lim _{x \rightarrow 0}\left(\frac{1+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)^{\frac{1}{\sin ^{3} x}}=$

$=\lim _{x \rightarrow 0}\left(e^{\ln \left(\left(1+x^{2} \cdot 2^{x}\right) /\left(1+x^{2} \cdot 5^{x}\right)\right)}\right)^{\frac{1}{\sin ^{3} x}}=$

$=\lim _{x \rightarrow 0} e^{\frac{1}{\sin ^{3} x} \ln \left(\left(1+x^{2} \cdot 2^{x}\right) /\left(1+x^{2} \cdot 5^{x}\right)\right)}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left(\left(1+x^{2} \cdot 2^{x}\right) /\left(1+x^{2} \cdot 5^{x}\right)\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left(\frac{1+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left(\frac{1+x^{2} \cdot 5^{x}-x^{2} \cdot 5^{x}+x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left(1-\frac{x^{2} \cdot 5^{x}-x^{2} \cdot 2^{x}}{1+x^{2} \cdot 5^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{\sin ^{3} x} \ln \left(1-x^{2} \cdot \frac{5^{x}-2^{x}}{1+x^{2} \cdot 5^{x}}\right)\right\}=$

Using the substitution of equivalent infinitesimals:
$\ln \left(1-x^{2} \cdot \frac{5^{x}-2^{x}}{1+x \cdot 5^{x}}\right) \sim -x^{2} \cdot \frac{5^{x}-2^{x}}{1+x \cdot 5^{x}}$, as $x \rightarrow 0\left(-x^{2} \cdot \frac{5^{x}-2^{x}}{1+x \cdot 5^{x}} \rightarrow 0\right)$ $\sin x \sim x$, as $x \rightarrow 0$

We get:

$=\exp \left\{\lim _{x \rightarrow 0} \frac{1}{x^{3}}\left(-x^{2} \cdot \frac{5^{x}-2^{x}}{1+x^{2} \cdot 5^{x}}\right)\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{5^{x}-2^{x}}{x \cdot\left(1+x \cdot 5^{x}\right)}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{\left(5^{x}-1\right)-\left(2^{x}-1\right)}{x \cdot\left(1+x \cdot 5^{x}\right)}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{5^{x}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}+\lim _{x \rightarrow 0} \frac{2^{x}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{\left(e^{\ln 5}\right)^{x}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}+\lim _{x \rightarrow 0} \frac{\left(e^{\ln 2}\right)^{x}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}\right\}=$

$=\exp \left\{\lim _{x \rightarrow 0}-\frac{e^{x \cdot \ln 5}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}+\lim _{x \rightarrow 0} \frac{e^{x \cdot \ln 2}-1}{x \cdot\left(1+x \cdot 5^{x}\right)}\right\}=$

Using the substitution of equivalent infinitesimals:

$e^{x \cdot \ln 5}-1 \sim x \cdot \ln 5$, as $x \rightarrow 0(x \cdot \ln 5 \rightarrow 0)$

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

We get:

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

## Problem Kuznetsov Limits 17-31