Question

Solve :
$\underset{z\to 4}{lim}\frac{\sqrt{z}-2}{z-4}$

Answer 1

$\begin{array}{c}{lim}_{z\to 4}\frac{\sqrt{z}-2}{z-4}\\ ={lim}_{z\to 4}\frac{\left(\sqrt{z}-2\right)}{\left(z-4\right)}\frac{\left(\sqrt{z}+2\right)}{\left(\sqrt{z}+2\right)}\\ ={lim}_{z\to 4}\frac{z-4}{\left(z-4\right)\left(\sqrt{z}+2\right)}\\ ={lim}_{z\to 4}\frac{1}{\sqrt{z}+2}\\ =\frac{1}{4}\end{array}$