## Task Condition

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

$$
f(x)=\left\{\begin{array}{c}
\sin x \cdot \cos \frac{5}{x}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\sin \Delta x \cdot \cos \frac{5}{\Delta x}-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\sin \Delta x \cdot \cos \frac{5}{\Delta x}}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

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

We get:

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

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

## Problem Kuznetsov Differentiation 2-10