Question

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

localid="1647002262582" $$\begin{gathered} f\left(x\right)=\frac{x^3-8}{x^2+4} \\ c=2 \end{gathered}$$

Short answer

The function is continuous at$c$

Step-by-step solution

Given information

Given function

$$\begin{gathered} f\left(x\right)=\frac{x^3-8}{x^2+4} \\ c=2 \end{gathered}$$

Check whether the function is defined at c

We have

$$\begin{gathered} f\left(2\right)=\frac{2^3-8}{2^2+4} \\ f\left(2\right)=0 \end{gathered}$$

So it is defined at$c$

Check the left hand limits and right hand limits

Checking, we get

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{2^3-8}{2^2+4} \\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }\frac{2^3-8}{2^2+4} \\ \underset{x\to 2^{-}}{\lim }f\left(x\right)=0 \end{gathered}$$

Again,

$$\begin{gathered} \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{+}}{\lim }\frac{2^3-8}{-2^2+4} \\ \underset{x\to 2^{+}}{\lim }f\left(x\right)=0 \end{gathered}$$

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