Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 10

Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at the given number $a = 4$ using the definition of continuity and the properties of limits.

By using the properties of limit,


$\begin{equation} \begin{aligned} & \lim \limits_{x \to 4} (x^2 + \sqrt{7 -x}) && = \lim \limits_{x \to 4} x^2 + \sqrt{\lim \limits_{x \to 4} 7 - \lim \limits_{x \to 4} x} && \text{ Apply sum, difference}\\ & \phantom{x} && = (4)^2 + \sqrt{7-4} && \text{ Substitute the given value}\\ & \phantom{x} && = 16 + \sqrt{3} \end{aligned} \end{equation}$


By using the definition of continuity,
$\lim \limits_{x \to a} f(x) = f(a)$


$\begin{equation} \begin{aligned} & \lim \limits_{x \to 4} (x^2+ \sqrt{7-x})&& = f(4) = (4)^2 + \sqrt{7-4}\\ & \phantom{x} && = 16 + \sqrt{3} \end{aligned} \end{equation}$




Therefore, by applying either of the two, we have shown that the function is continuous at 4 and is equal to $16 + \sqrt{3}$