## Problem Statement

Calculate the limit of the function:

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

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

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

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

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

Using the substitution of equivalent infinitesimals:

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

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

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

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

We get:

$=\frac{\lim _{x \rightarrow 0} \frac{5 x}{x}-\lim _{x \rightarrow 0} \frac{3 x}{x}}{\lim _{x \rightarrow 0} \frac{2 x}{x}-\lim _{x \rightarrow 0} \frac{x}{x}}=\frac{\lim _{x \rightarrow 0} 5-\lim _{x \rightarrow 0} 3}{\lim _{x \rightarrow 0} 2-\lim _{x \rightarrow 0} 1}=\frac{5-3}{2-1}=2$