Chapter 1.1, Problem 60E

(a)

To determine

To draw: The graph of $f$ .

Answer to Problem 60E

The graph is shown:

  

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=-1,\\ f(x)=\left\{\begin{array}{c}1-x^2,x\ne -1\\ 2,x=-1\end{array}\right.\end{array}$

Calculation:

The graph is shown below:

  

(b)

To determine

To find: The value of $\underset{x\to c^{+}}{\lim }f(x)$ and

  $\underset{x\to c^{-}}{\lim }f(x)$ .

Answer to Problem 60E

All of $x$ .

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=-1,\\ f(x)=\left\{\begin{array}{c}1-x^2,x\ne -1\\ 2,x=-1\end{array}\right.\end{array}$

Calculation:

The limit exists anywhere where the function is continuous, that is, on the interval $(-\infty ,-1)\cup (-1,\infty )$ .

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

and as both limits exist and are equal, then $\underset{x\to -1}{\lim }f(x)$ exists as well.

Therefore the required the limit exists for all real $x$ .

(c)

To determine

To find: The limit $\underset{x\to c}{\lim }f(x)$ exists or no.

Answer to Problem 60E

  $\underset{x\to -1}{\lim }f(x)=0$

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=-1,\\ f(x)=\left\{\begin{array}{c}1-x^2,x\ne -1\\ 2,x=-1\end{array}\right.\end{array}$

Calculation:

From part (b):

  $\underset{x\to -1^{+}}{\lim }f(x)=0$ and

  $\underset{x\to -1^{-}}{\lim }f(x)=0$

Therefore, $\underset{x\to -1}{\lim }f(x)=0$ . Hence the left- and right-hand limits exist and are equal.

Therefore, the required $\underset{x\to -1}{\lim }f(x)=0$ .