## Problem Statement

Calculate the limit of the function:

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

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

Using the substitution of equivalent infinitesimals:

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

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

$\sin 3 x \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$

$\tan 2 x \sim 2 x$, as $x \rightarrow 0(2 x \rightarrow 0)$

We get:

$$
=\frac{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}}{\lim _{x \rightarrow 0} \frac{3 x}{x}-\lim _{x \rightarrow 0} \frac{2 x}{x}}=\frac{\lim _{x \rightarrow 0} 3-\lim _{x \rightarrow 0} 2}{\lim _{x \rightarrow 0} 3-\lim _{x \rightarrow 0} 2}=\frac{3-2}{3-2}=1
$$