Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 59

Give an example that $\lim \limits_{x \to a} [f(x) g(x)]$ may exist even though neither $\lim \limits_{x \to a} f(x)$ nor $\lim \limits_{x \to a} g(x)$ exists.

Suppose that $\displaystyle f(x) \frac{x - 4}{\sqrt{x}}$ and $\displaystyle g(x) = \frac{x^2 - 2x}{\sqrt{x}}$


$\begin{equation} \begin{aligned} & \lim \limits_{x \to 0} f(x) \text{ and } \lim \limits_{x \to 0} g(x) \text{ do not exists for the functions are not defined for $$ denominator}\\ \\ \text{ But, } & \lim \limits_{x \to 0} [f(x) g(x)] = \lim \limits_{x \to 0} \left[ \left( \frac{x - 4}{\sqrt{x}} \right) \left( \frac{x^2 - 2x}{\sqrt{x}} \right) \right] \qquad = \lim \limits_{x \to 0}\left[\frac{(x-4)(x^2-2x)}{(\sqrt{x})(\sqrt{x})}\right]\\ \\ & \lim \limits_{x \to 0} [f(x)g(x)] = \lim \limits_{x \to 0} \left[ \frac{x^3 - 4x^2 - 2x^2 + 8x}{x} \right]\\ \\ & \lim \limits_{x \to 0} [f(x)g(x)] = \lim \limits_{x \to 0} \left[ \frac{\cancel{x}(x^2 - 6x + 8)}{\cancel{x}} \right] \qquad = \lim \limits_{x \to 0} (x^2-6x+8)\\ \\ & \lim \limits_{x \to 0} [f(x)g(x)] = (0)^2 - 6(0) + 8 = 8\\ \\ & \lim \limits_{x \to 0} [f(x)g(x)] = 8 \end{aligned} \end{equation}$