Answer
$\lim\limits_{x\to2}x\sqrt{20-x^2}=8$
Work Step by Step
$$\lim\limits_{x\to2}x\sqrt{20-x^2}=\lim\limits_{x\to2}f(x)$$
1) Determine whether $f(x)$ is continuous at $2$
$f(x)$ is defined where $$(20-x^2)\ge0$$$$x^2\le20$$$$-\sqrt{20}\le x\le\sqrt{20}$$
Therefore, the domain of $f(x)$ is $[-\sqrt{20},\sqrt{20}]$.
By Theorem 7, $f(x)$ is continuous on its domain: $[-\sqrt{20},\sqrt{20}]$.
Since $2\in[-\sqrt{20},\sqrt{20}]$, $f(x)$ is also continuous at $2$.
2) By definition, $f(x)$ is continuous at $2$ if and only if $$\lim\limits_{x\to2}f(x)=f(2)$$$$\lim\limits_{x\to2}x\sqrt{20-x^2}=2\times\sqrt{20-2^2}=8$$
