To be calculated $\lim _{x \rightarrow 0} \frac{1-\sqrt[3]{1-x}}{x}$.
For the indeterminate quotient, as a function of $x$, the substitution value at $x=0$ is $\frac{0}{0}$. Therefore, at $x=0$, we can only speak of a limit.

# Let 

$$
\sqrt[3]{1-x}=y, \quad \text{so} \quad x=1-y^{3}
$$

The fractional expression: $\frac{1-y}{1-y^{3}}=\frac{1}{1+y+y^{2}}$ can be written. The value of the fractional expression $\frac{1-y}{1-y^{3}}$ is always given by the value of $\frac{1}{1+y+y^{2}}$; if $y=1$, then we consider it as a limit.

Already now, if $x \rightarrow 0$, then $y \rightarrow 1$ and thus the limit of the expression is $\frac{1}{3}$.

Tölösi Márta and Varga Irén (Érseki Girls' High School VII. grade, Esztergom)