Determining Whether a Function Is Continuous on a Closed Interval

Is the function \(f(x)=\sqrt{4-x^{2}}\) continuous on the closed interval \([-2,2]\)?

Solution The domain of \(f\) is \(\{x|-2\leq x\leq 2\}\). So, \(f\) is defined for every number in the closed interval \([-2,2]\).

For any number \(c\) in the open interval \((-2,2)\), \begin{equation*} \lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}\sqrt{4-x^{2}}=\sqrt{ \lim\limits_{x\rightarrow c}(4-x^{2})}=\sqrt{4-c^{2}}=f(c) \end{equation*}

So, \(f\) is continuous on the open interval \((-2,2)\).

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To determine whether \(f\) is continuous on \([-2,2]\), we investigate the limit from the right at \(-2\) and the limit from the left at \(2\). Then, \begin{equation*} \lim_{x\rightarrow -2^{+}}f(x)=\lim_{x\rightarrow -2^{+}}\sqrt{4-x^{2}} =0=f(-2) \end{equation*}

So, \(f\) is continuous from the right at \(-2\). Similarly, \begin{equation*} \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{-}}\sqrt{4-x^{2}} =0=f(2) \end{equation*}

So, \(f\) is continuous from the left at \(2\). We conclude that \(f\) is continuous on the closed interval \([-2,2]\).