Example 13. Find $\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{\sqrt{1+x}-1}$.
Solution. Introduce a new variable $z$ by the formula

$$
1+x=z^{6}
$$

(the exponent is chosen so that it is possible to extract both a square and a cube root).

From equality (A), it follows that $z \rightarrow 1$ as $x \rightarrow 0$. Substituting the expression of the old variable in terms of the new one, factoring the numerator and the denominator, we get

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-1}{\sqrt{1+x}-1}=\lim _{z \rightarrow 1} \frac{\sqrt[3]{z^{6}}-1}{\sqrt{z^{6}}-1}=\lim _{z \rightarrow 1} \frac{z^{2}-1}{z^{3}-1}= \\
= & \lim _{z \rightarrow 1} \frac{(z-1)(z+1)}{(z-1)\left(z^{2}+z+1\right)}=\lim _{z \rightarrow 1} \frac{z+1}{z^{2}+z+1}=\frac{2}{3}
\end{aligned}
$$