## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{5^{2 x}-2^{3 x}}{\sin x+\sin x^{2}}$
## Solution

$\lim _{x \rightarrow 0} \frac{5^{2 x}-2^{3 x}}{\sin x+\sin x^{2}}=\lim _{x \rightarrow 0} \frac{\left(25^{x}-1\right)-\left(8^{x}-1\right)}{\sin x+\sin x^{2}}=$

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 25}\right)^{x}-1\right)-\left(\left(e^{\ln 8}\right)^{x}-1\right)}{\sin x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 25}-1\right)-\left(e^{x \ln 8}-1\right)}{\sin x+\sin x^{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 25}-1\right)-\left(e^{x \ln 8}-1\right)\right)}{\frac{1}{x}\left(\sin x+\sin x^{2}\right)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 25}-1\right)-\left(e^{x \ln 8}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\sin x+\sin x^{2}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 25}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 8}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\sin x}{x}+\lim _{x \rightarrow 0} \frac{\sin x^{2}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 25}-1 \sim x \ln 25$, as $x \rightarrow 0(x \ln 25 \rightarrow 0)$

$e^{x \ln 8}-1 \sim x \ln 8$, as $x \rightarrow 0(x \ln 8 \rightarrow 0)$

$\sin x \sim x$, as $x \rightarrow 0$

$\sin x^{2} \sim x^{2}$, as $x \rightarrow 0\left(x^{2} \rightarrow 0\right)$

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{x \ln 25}{x}-\lim _{x \rightarrow 0} \frac{x \ln 8}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{x^{2}}{x}}=\frac{\lim _{x \rightarrow 0} \ln 25-\lim _{x \rightarrow 0} \ln 8}{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0} x}=$
$=\frac{\ln 25-\ln 8}{1+0}=\ln \frac{25}{8}$

## Problem Kuznetsov Limits 15-23