Use the definition of continuity and the properties of limits to show that the f

Use the definition of continuity and the properties of limits to show that the function
$f\left(x\right)=x\sqrt{16-{x}^{2}}$ is continuous on the interval [-4,4]
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Nathaniel Kramer
Consider the given function as
$f\left(x\right)=x\sqrt{16-{x}^{2}}$
Definition of continuity
Function is said to be continuous if $\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=f\left(a\right)$
Let a point $c\in \left[-4,4\right]$ then,
$\underset{x\to {c}^{-}}{lim}f\left(x\right)=\underset{x\to {c}^{-}}{lim}x\sqrt{16-{x}^{2}}$
$=c\sqrt{16-{c}^{2}}$
And
$\underset{x\to {c}^{+}}{lim}f\left(x\right)=\underset{x\to {c}^{+}}{lim}x\sqrt{16-{x}^{2}}$
$=c\sqrt{16-{c}^{2}}$
And
$f\left(c\right)=c\sqrt{16-{c}^{2}}$
So, $\underset{x\to {c}^{-}}{lim}f\left(x\right)=\underset{x\to {c}^{+}}{lim}f\left(x\right)=f\left(c\right)$
Hence, function is continuous on the given interval [-4,4]
Jeffrey Jordon