16 December 2021

Statement

Is the following function continuous on the interval \(x \in [-4, 4]\) ?

\(f(x) = x \sqrt{16 - x^2}\)

Solution

\(f(x) = x \sqrt{16 - x^2}\)

Let's start by finding the limit when \(x \to a\) with \(-4 < a < 4\) .


\(\lim\limits_{x \to a} f(x)\)

\(= \lim\limits_{x \to a} x \sqrt{16 - x^2}\)

\(= \lim\limits_{x \to a} x \times \lim\limits_{x \to a} \sqrt{16 - x^2}\)

\(= \lim\limits_{x \to a} x \times \sqrt{\lim\limits_{x \to a} (16 - x^2)}\)

\(= a \sqrt{16 - a^2}\)

\(= f(a)\)

If \(|a| > 4\) then the value of \(f(a)\) becomes undefined in \(\mathbb{R}\) but since \(-4 < a < 4\) the function is continous in that interval. The last to step is to find \( \lim\limits_{x \to -4^{+}}\) and \( \lim\limits_{x \to 4^{-}}\).


\( \lim\limits_{x \to -4^{+}} f(x)\)

\(= \lim\limits_{x \to -4^{+}} x \sqrt{16 - x^2}\)

\(= 4 \sqrt{16 - (-4)^2}\)

\(= 0\)

\(= f(-4)\)


Finally we check the right bound of the interval.

\( \lim\limits_{x \to 4^{-}} f(x)\)

\(= \lim\limits_{x \to 4^{-}} x \sqrt{16 - x^2}\)

\(= 4 \sqrt{16 - 4^2}\)

\(= 0\)

\(= f(4)\)


Therefore \(f(x)\) is continuous when \(x \in [-4, 4]\).