# Evaluate the following limits  $\lim\limits_{\large z \to 1} \large\frac{\large z^{\large\frac{1}{3}}-1}{z^{\large\frac{1}{6}}-1}$

$\begin{array}{1 1}2 \\ 3\\ 4\\ 0 \end{array}$

$\lim\limits_{z \to 1} \large\frac{z^{\Large\frac{1}{3}}-1}{z^{\Large\frac{1}{6}}-1}$
At $z = 1$ the value of the given fraction takes the form $\large\frac{0}{0}.$
Put $z^{\large\frac{1}{6}}=x$ so that $z \rightarrow 1$ as $x \rightarrow 1$.
Accordingly, $\lim\limits_{z \to 1} \large\frac{z^{\Large\frac{1}{3}}-1}{z^{\Large\frac{1}{6}}-1}$ $= \lim\limits_{z \to 1} \large\frac{x^2-1}{x-1}$
$= \lim\limits_{z \to 1} \large\frac{x^2-1^2}{x-1}$
$= 2.1^{2-1} \quad \quad \quad \bigg[ \lim\limits_{x \to a} \large\frac{x^n-a^n}{x-a} = na^{n-1} \bigg]$
$= 2$