## Problem Statement

Based on the definition of the derivative, find $f^{\prime}(0)$:

$$
f(x)=\left\{\begin{array}{c}
e^{x \sin \frac{5}{x}}-1, x \neq 0 \\
0, x=0
\end{array}\right.
$$
## Solution

By definition, the derivative at the point $x=0$:

$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$

Based on the definition, we find:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(e^{\Delta x \sin \frac{5}{\Delta x}}-1-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0}\left(e^{\Delta x \sin \frac{5}{\Delta x}}-1\right) / \Delta x=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
e^{\Delta x \sin \frac{5}{\Delta x}}-1 \sim \Delta x \sin \frac{5}{\Delta x}_{\text {, as }} \Delta x \rightarrow 0\left(\Delta x \sin \frac{5}{\Delta x} \rightarrow 0\right)
$$

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{\Delta x \sin \frac{5}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0} \sin \frac{5}{\Delta x} \text { - the limit does not exist. }
$$

Thus, $f^{\prime}(0)$ - does not exist.

## Problem Kuznetsov Differentiation 1-20(2)