Chapter 14.3, Problem 47AYU

$f\left(x\right)=\frac{x^2+5}{x-6}c=3$

To determine

To find: Whether $f$ is continuous at $c$.

Answer to Problem 47AYU

Continuous.

Explanation of Solution

Given:

$f\left(x\right)=\frac{x^2+5}{x-6}$; $c=3$

Calculation:

We see that,

$\underset{x\to c}{\lim }f\left(x\right)=\underset{x\to 3}{\lim }\left(\frac{x^2+5}{x-6}\right)=\left(\frac{3^2+5}{3-6}\right)=\left(\frac{9+5}{-3}\right)=-\frac{14}{3}$

Again,

$f\left(c\right)=f\left(3\right)=\frac{3^2+5}{3-6}=\frac{9+5}{-3}=-\frac{14}{3}$

Since $\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$; therefore $f$ is continuous at $c$.