Answer
$$
\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{2}{3}
$$
Work Step by Step
$$
\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}
$$
Let $t=\sqrt[6]{x},$ so $x=t^{6} .$ Then $t \rightarrow 1$ as $x \rightarrow 1,$ so
$$
\begin{aligned}
\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1}&=\lim _{t \rightarrow 1} \frac{t^{2}-1}{t^{3}-1} \\
&=\lim _{t \rightarrow 1} \frac{(t-1)(t+1)}{(t-1)\left(t^{2}+t+1\right)}\\
&=\lim _{t \rightarrow 1} \frac{t+1}{t^{2}+t+1}\\
&=\frac{1+1}{1^{2}+1+1}\\
&=\frac{2}{3}
\end{aligned}
$$
