## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{7^{3 x}-3^{2 x}}{\tan x+x^{3}}$
## Solution

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{7^{3 x}-3^{2 x}}{\tan x+x^{3}}=\lim _{x \rightarrow 0} \frac{\left(343^{x}-1\right)-\left(9^{x}-1\right)}{\tan x+x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\left(e^{\ln 343}\right)^{x}-1\right)-\left(\left(e^{\ln 9}\right)^{x}-1\right)}{\tan x+x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(e^{x \ln 343}-1\right)-\left(e^{x \ln 9}-1\right)}{\tan x+x^{3}}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x \ln 343}-1\right)-\left(e^{x \ln 9}-1\right)\right)}{\frac{1}{x}\left(\tan x+x^{3}\right)}= \\
& =\frac{\lim _{x \rightarrow 0} \frac{1}{x}\left(\left(e^{x \ln 343}-1\right)-\left(e^{x \ln 9}-1\right)\right)}{\lim _{x \rightarrow 0} \frac{1}{x}\left(\tan x+x^{3}\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{x \ln 343}-1}{x}-\lim _{x \rightarrow 0} \frac{e^{x \ln 9}-1}{x}\right) /\left(\lim _{x \rightarrow 0} \frac{\tan x}{x}+\lim _{x \rightarrow 0} \frac{x^{3}}{x}\right)=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{x \ln 343}-1 \sim x \ln 343$, as $x \rightarrow 0(x \ln 343 \rightarrow 0)$

$e^{x \ln 9}-1 \sim x \ln 9$, as $x \rightarrow 0(x \ln 9 \rightarrow 0)$

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

We get:

$$
\begin{aligned}
& =\frac{\lim _{x \rightarrow 0} \frac{x \ln 343}{x}-\lim _{x \rightarrow 0} \frac{x \ln 9}{x}}{\lim _{x \rightarrow 0} \frac{x}{x}+\lim _{x \rightarrow 0} \frac{x^{3}}{x}}=\frac{\lim _{x \rightarrow 0} \ln 343-\lim _{x \rightarrow 0} \ln 9}{\lim _{x \rightarrow 0} 1+\lim _{x \rightarrow 0} x^{2}}= \\
& =\frac{\ln 343-\ln 9}{1+0}=\ln \frac{343}{9}=\ln \frac{7^{3}}{3^{2}}
\end{aligned}
$$

Problem Kuznetsov Limits 15-17