Chapter 2, Problem 29RE

(a)

To determine

To find: The right-hand and left-hand limits of the given function at $x=-1$ , $x=0$ and $x=1$ .

Answer to Problem 29RE

The right-hand and left-hand limits both at $x=-1$ is 1, at $x=0$ is 0. The right-hand limit at $x=1$ is 1 and left-hand limit is $-1$ .

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}1,x\le -1\\ -x,-1<x<0\\ 1,x=0\\ -x,0<x<1\\ 1,x\ge 1\end{array}$ .

Calculation:

The left-hand limit at $x=-1$ is:

  $\begin{array}{c}{\lim }_{x\to -1^{-}}f\left(x\right)={\lim }_{x\to -1^{-}}\left(1\right)\\ =1\end{array}$

The right-hand limit at $x=-1$ is:

  $\begin{array}{c}{\lim }_{x\to -1^{+}}f\left(x\right)={\lim }_{x\to -1^{+}}\left(-x\right)\\ =-\left(-1\right)\\ =1\end{array}$

The left-hand limit at $x=0$ is:

  $\begin{array}{c}{\lim }_{x\to 0^{-}}f\left(x\right)={\lim }_{x\to 0^{-}}\left(-x\right)\\ =0\end{array}$

The right-hand limit at $x=0$ is:

  $\begin{array}{c}{\lim }_{x\to 0^{+}}f\left(x\right)={\lim }_{x\to 0^{+}}\left(-x\right)\\ =0\end{array}$

The left-hand limit at $x=1$ is:

  $\begin{array}{c}{\lim }_{x\to 1^{-}}f\left(x\right)={\lim }_{x\to 1^{-}}\left(-x\right)\\ =-1\end{array}$

The right-hand limit at $x=1$ is:

  $\begin{array}{c}{\lim }_{x\to 1^{+}}f\left(x\right)={\lim }_{x\to 1^{+}}\left(1\right)\\ =1\end{array}$

Therefore, the right-hand and left-hand limits both at $x=-1$ is 1, at $x=0$ is 0. The right-hand limit at $x=1$ is 1 and left-hand limit is $-1$ .

(b)

To determine

To find:The function has limits if $x$ approachesto $x=-1$ , $x=0$ and $x=1$ .

Answer to Problem 29RE

The limit at $x=-1$ is 1, the limit at $x=0$ is 0 and the limit at $x=1$ does not exist as left-hand limit is not equal to the right-hand limit.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}1,x\le -1\\ -x,-1<x<0\\ 1,x=0\\ -x,0<x<1\\ 1,x\ge 1\end{array}$ .

Calculation:

From part (a), observed the limits of the function at $x=-1$ , $x=0$ and $x=1$ .

At $x=-1$ , the right-hand limit of the function is equal to the left-hand limit. So, the limit exists at this point.

  $\underset{x\to -1}{\lim }f\left(x\right)=1$

At $x=0$ , the right-hand limit of the function is equal to the left-hand limit. So, the limit exists at this point.

  $\underset{x\to 0}{\lim }f\left(x\right)=0$

At $x=1$ , the right-hand limit of the function is not equal to the left-hand limit of the function. So, the limit does not exist at this point.

Therefore, the limit at $x=-1$ is 1, the limit at $x=0$ is 0 and the limit at $x=1$ does not exist as left-hand limit is not equal to the right-hand limit.

(c)

To determine

To find: The given function is continuous at points $x=-1$ , $x=0$ and $x=1$ .

Answer to Problem 29RE

The function is continuous at $x=-1$ as the value of the function and limit is equal.

Explanation of Solution

Given information:

The given function is $f\left(x\right)=\{\begin{array}{l}1,x\le -1\\ -x,-1<x<0\\ 1,x=0\\ -x,0<x<1\\ 1,x\ge 1\end{array}$ .

Calculation:

If the limit of a function is equal to the value of the function at that point, then the function is said to be continuous.

The limit at $x=-1$ is 1. The value of the function at $x=-1$ is,

  $f\left(-1\right)=1$

So, the function is continuous at $x=-1$ .

The limit at $x=0$ is 0. The value of the function at $x=0$ is,

  $f\left(0\right)=1$

So, the function is discontinuous at $x=0$ .

The limit does not exist at $x=1$ . So, function is discontinuous.

Therefore, the function is continuous at $x=-1$ as the value of the function and limit is equal.