Question

$\underset{x\to 4}{\lim }\frac{2-\sqrt{x}}{4-x}$

Answer 1

$\underset{x\to 4}{\lim }\left[\frac{2-\sqrt{x}}{4-x}\right]$

= $\underset{x\to 4}{\lim }\left[\frac{2-\sqrt{x}}{2-{\left(\sqrt{x}\right)}^2}\right]$

= $\underset{x\to 4}{\lim }\left[\frac{\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right]$

= $\frac{1}{2+\sqrt{4}}$

= $\frac{1}{2+2}$

= $\frac{1}{4}$