## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}$
## Solution

$\lim _{x \rightarrow 0} \frac{e^{\alpha x}-e^{\beta x}}{\sin \alpha x-\sin \beta x}=\lim _{x \rightarrow 0} \frac{\left(e^{\alpha x}-1\right)-\left(e^{\beta x}-1\right)}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$
$=\lim _{x \rightarrow 0} \frac{e^{\alpha x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{e^{\beta x}-1}{2 \sin \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}=$

Using the substitution of equivalent infinitesimals:

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

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

$\sin \frac{x(\alpha-\beta)}{2} \sim \frac{x(\alpha-\beta)}{2}, \quad x \rightarrow 0\left(\frac{x(\alpha-\beta)}{2} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\alpha x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta x}{2 \frac{x(\alpha-\beta)}{2} \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\lim _{x \rightarrow 0} \frac{\alpha}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}-\lim _{x \rightarrow 0} \frac{\beta}{(\alpha-\beta) \cos \frac{x(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}-\frac{\beta}{(\alpha-\beta) \cos \frac{0 \cdot(\alpha+\beta)}{2}}= \\
& =\frac{\alpha}{(\alpha-\beta) \cos 0}-\frac{\alpha}{(\alpha-\beta) \cos 0}=\frac{\alpha}{(\alpha-\beta) \cdot 1}-\frac{\beta}{(\alpha-\beta) \cdot 1}= \\
& =\frac{\alpha-\beta}{\alpha-\beta}=1
\end{aligned}
$$

## Problem Kuznetsov Limits 16-6