Evaluate the Given limit:

$\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

At $z=1$, the value of the given function takes the form $\frac{0}{0}$.

Put $z^{\frac{1}{6}}=x$ so that $z \rightarrow 1$ as $x \rightarrow 1$.

Accordingly, $\lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$

$=\lim _{x \rightarrow 1} \frac{x^{2}-1^{2}}{x-1}$

$=2 \cdot 1^{2-1}$ $\left[\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}\right]$

$=2$

$\therefore \lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$