Chapter 3.1, Problem 10QR

To determine

Whether the function $f$ is continuous or not.

Answer to Problem 10QR

The function f is not continuous.

Explanation of Solution

Given information:

The function: $f\left(x\right)=\{\begin{array}{cc}x+2,& x\le 1\\ {\left(x-1\right)}^2& x>1\end{array}$

The given functions are,

  $f\left(x\right)=x+2$ , $x\le 1$

And, $f\left(x\right)={\left(x-1\right)}^2$ , $x>1$

For $x\to 1^{+}$ ; we can use the function $f\left(x\right)={\left(x-1\right)}^2$ .

So, $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }{\left(x-1\right)}^2\\ ={\left(1-1\right)}^2\\ =0\end{array}$

For $x\to 1^{-}$ , we can use the function $f\left(x\right)=x+2$ .

So, $\begin{array}{c}\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(x+2\right)\\ =1+2\\ =3\end{array}$

Here, $\underset{x\to 1^{+}}{\lim }f\left(x\right)\ne \underset{x\to 1^{-}}{\lim }f\left(x\right)$

So, the limit $\underset{x\to 1}{\lim }f\left(x\right)$ does not exist.

Now, the limit of the function must exist for the function f to be continuous.

But, the limit of the function f does not exist.

Hence, the function f is not continuous.