Answer
$\lim\limits_{x \to 4}\frac{4x-x^{2}}{2-\sqrt x} = 16$
Work Step by Step
$\frac{4x-x^{2}}{2-\sqrt x} = \frac{x(4-x)}{2-\sqrt x} = \frac{x(2-\sqrt x)(2+\sqrt x)}{2-\sqrt x} = x(2+\sqrt x)$
Now,
$\lim\limits_{x \to 4}\frac{4x-x^{2}}{2-\sqrt x} = \lim\limits_{x \to 4}x(2+\sqrt x) = 4(2+2)= 16$
