Answer
16
Work Step by Step
We want to find $\lim\limits_{x \to 4}\dfrac{4x-x^2}{2-\sqrt{x}}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 4 is zero, and if we try to substitute $x=4$ directly, we get zero in both the numerator and the denominator.
But note that $\dfrac{4x-x^2}{2-\sqrt{x}}=\dfrac{x(4-x)}{2-\sqrt{x}}=\dfrac{x(2-\sqrt{x})(2+\sqrt{x})}{2-\sqrt{x}}=x(2+\sqrt{x})$
for all $x \neq 4.$
Thus $\lim\limits_{x \to 4}\dfrac{4x-x^2}{2-\sqrt{x}}=\lim\limits_{x \to 4}x(2+\sqrt{x})=4(2+\sqrt{4})=4(2+2)=4(4)=16.$
