## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to2}x\sqrt{20-x^2}=8$
$$\lim\limits_{x\to2}x\sqrt{20-x^2}=\lim\limits_{x\to2}f(x)$$ 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 $$\lim\limits_{x\to2}f(x)=f(2)$$$$\lim\limits_{x\to2}x\sqrt{20-x^2}=2\times\sqrt{20-2^2}=8$$