## Problem Statement

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

$$
f(x)=\left\{\begin{array}{c}
\sin \left(x \sin \frac{3}{x}\right), x \neq 0 \\
0, x=0
\end{array}\right.
$$
## 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:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \frac{\sin \left(\Delta x \cdot \sin \frac{3}{\Delta x}\right)-0}{\Delta x}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{\sin \left(\Delta x \cdot \sin \frac{3}{\Delta x}\right)}{\Delta x}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$$
\sin \left(\Delta x \cdot \sin \frac{3}{\Delta x}\right) \sim \Delta x \cdot \sin \frac{3}{\Delta x}_{, \text {as }} \Delta x \rightarrow 0\left(\Delta x \cdot \sin \frac{3}{\Delta x} \rightarrow 0\right)
$$

We get:

$$
=\lim _{\Delta x \rightarrow 0} \frac{\Delta x \cdot \sin \frac{3}{\Delta x}}{\Delta x}=\lim _{\Delta x \rightarrow 0} \sin \frac{3}{\Delta x} \text { - the limit does not exist. }
$$

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

## Problem Kuznetsov Differentiation $2-5$