Chapter 1.1, Problem 59E

(a)

To determine

To draw: The graph of $f$ .

Answer to Problem 59E

The graph is shown:

  

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=1,\\ f(x)=\left\{\begin{array}{c}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}\right.\end{array}$

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 59E

  $\underset{x\to 1^{+}}{\lim }f(x)=4$ and

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

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=1,\\ f(x)=\left\{\begin{array}{c}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}\right.\end{array}$

Calculation:

Put, $x=1$ and calculate the limit:

  $\begin{array}{c}\underset{x\to 1^{+}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }{x}^3-2x+5\\ =1^3-2\cdot 1+5\\ =4\end{array}$

The function $\frac{1}{x-1}$ and calculate the limit of $x=1$ :

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

Therefore the required value of $\underset{x\to c^{+}}{\lim }f(x)$ and $\underset{x\to c^{-}}{\lim }f(x)$ are:

  $\underset{x\to 1^{+}}{\lim }f(x)=4$ and

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

(c)

To determine

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

Answer to Problem 59E

Does not exists.

Explanation of Solution

Given information: The functions are:

  $\begin{array}{c}c=1,\\ f(x)=\left\{\begin{array}{c}\frac{1}{x-1},x<1\\ x^3-2x+5,x\ge 1\end{array}\right.\end{array}$

Calculation:

From part (b) $\underset{x\to 1^{+}}{\lim }f(x)=2$ and

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

For the limit to exist, the right- and left-hand limits must both exist and be equal. Since the left-hand limit does not exist $\underset{x\to 1}{\lim }f(x)$ does not exist.

Therefore the required $\underset{x\to 1}{\lim }f(x)$ does not exist.