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 $$x \in [-4, 4]$$.