Question

$\underset{z\to 1}{lim}\frac{z^{1_{/3}}-1}{z^{1_{/6}}-1}$.

Answer 1

$\underset{z\to 1}{lim}\frac{z^{1/3}-1}{z^{1/b}-1}$

$\Rightarrow \frac{1^{1/3}-1}{1^{1/b}-1}$

$\Rightarrow \frac{0}{0}$

Since, it is in $\frac{0}{0}$ from we can solve by along the below theorem

$\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n{a}^{n-1}---\left(1\right)$

Hence

$\underset{z\to 1}{lim}\frac{z^{1/3}-1}{z^{1/b}-1}$

$\underset{z\to 1}{lim}\frac{z^{1/3}-1^{1/3}}{z^{1/b}-1^{1/b}}$

Multiplying and adding by $z=1$

$\underset{z\to 1}{lim}\frac{z^{1/3}-1^{1/3}}{z-1}-\underset{z\to 1}{lim}\frac{z^{1/b}-1^{1/b}}{z-1}---\left(2\right)$

from $\left(1\right)$ & $\left(2\right)$

$\to \frac{1}{3}÷\frac{1}{6}$

$=2$