Question

In Problems 45-60, determine whether $f$is continuous at $c$.

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

Short answer

The function is continuous at$c$

Step-by-step solution

Given information

Given function

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

Check whether the function is defined at c

Checking, we get

$$\begin{gathered} f\left(3\right)=\frac{3^2+5}{3-6} \\ f\left(3\right)=\frac{14}{-3} \\ f\left(3\right)=-\frac{14}{3} \end{gathered}$$

So it is defined at$c$

Check the left hand limits and right hand limits

Checking, we get

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

Again,

$$\begin{gathered} \underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }\frac{{\displaystyle 3^2+5}}{{\displaystyle 3-6}} \\ \underset{x\to 3^{+}}{\lim }f\left(x\right)=-\frac{14}{3} \end{gathered}$$

Since$\underset{x\to c}{\lim }f\left(x\right)=f\left(c\right)$so it is continuous at$c$