Question

Evaluate the given limit :
$\underset{z\to 1}{lim}\frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

Answer 1

Given : $\underset{z\to 1}{lim}\frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$
$=\frac{(1{)}^{\frac{1}{3}}-1}{(1{)}^{\frac{1}{6}}-1}$
$=\frac{1-1}{1-1}$
$=\frac{0}{0}$
Since it is form $\frac{0}{0}$

We simplify as,
$\underset{z\to 1}{lim}\frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=(\underset{x\to 1}{lim}{z}^{\frac{1}{3}}-1)÷(\underset{z\to 1}{lim}{z}^{\frac{1}{6}}-1)$
Multiplying and dividing by $z-1$
= $\underset{z\to 1}{lim}(z^{\frac{1}{3}}-1)\times \frac{(z-1)}{(z-1)}÷\underset{z\to 1}{lim}{z}^{\frac{1}{6}}-1$
= $\left({lim}_{z\to 1}\frac{z^{\frac{1}{3}}-1}{z-1}\right)÷\left({lim}_{z\to 1}\frac{z^{\frac{1}{6}}-1}{z-1}\right)$
= $\left(\underset{z\to 1}{lim}\frac{z^{\frac{1}{3}}-1^{\frac{1}{3}}}{z-1}\right)÷\left({lim}_{z\to 1}\frac{z^{\frac{1}{6}}-1^{\frac{1}{6}}}{z-1}\right)$

$\{\text{Using}\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}\}$
$=\frac{1}{3}(1{)}^{\frac{1}{3}-1}÷\frac{1}{6}(1{)}^{\frac{1}{6}-1}$
$=\frac{1}{3}÷\frac{1}{6}$
$=\frac{1}{3}\times \frac{6}{1}$
$=2$
$\therefore$ $\underset{z\to 1}{lim}\frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$