Chapter 11.3, Problem 81E

Solution

a)

To draw: The graph of $f\left(x\right)=\left\{\begin{array}{l}\sqrt{1-x^2}\text{ if -1}\le x<0\\ x\text{ if 0}\le x\text{<1}\\ \text{2 if }x=\text{1}\end{array}\right.$ .

For the given piecewise function the first piece is $y=\sqrt{1-x^2}$ with radius 1 centered at origin and second piece is $y=x$ which is a line with slope 1and third piece is point $\left(1,2\right)$ .

Given information: The given piecewise defined function is $f\left(x\right)=\left\{\begin{array}{l}\sqrt{1-x^2}\text{ if -1}\le x<0\\ x\text{ if 0}\le x\text{<1}\\ \text{2 if }x=\text{1}\end{array}\right.$

Calculation:

For the given piecewise function the first piece is $y=\sqrt{1-x^2}$ with radius 1 centered at origin which is the upper half of the circle and second piece is $y=x$ which is a line with slope 1and third piece is point $\left(1,2\right)$ .

Thus, the graph is plotted below:

  

b)

To find: The points $c$ in the domain of $f$where $\underset{x\to c}{\lim }f\left(x\right)$ exists.

The limit $\underset{x\to c}{\lim }f\left(x\right)$ exists and this occurs for all $x\in \left(-1,0\right)\cup \left(0,1\right)$ .

Given information: The given piecewise defined function is $f\left(x\right)=\left\{\begin{array}{l}\sqrt{1-x^2}\text{ if -1}\le x<0\\ x\text{ if 0}\le x\text{<1}\\ \text{2 if }x=\text{1}\end{array}\right.$

Calculation:

The limit $\underset{x\to c}{\lim }f\left(x\right)$ exists if and is equal to $L$ if and only if $\underset{x\to c^{-}}{\lim }f\left(x\right)=\underset{x\to c^{+}}{\lim }f\left(x\right)=L$ and this occurs for all $x\in \left(-1,0\right)\cup \left(0,1\right)$ .

c)

To find: The points $c$ where only the left-hand limit exists.

The left-hand limit of $\underset{x\to c}{\lim }f\left(x\right)$ at point $c$ exists at $x=1$ .

Given information: The given piecewise defined function is $f\left(x\right)=\left\{\begin{array}{l}\sqrt{1-x^2}\text{ if -1}\le x<0\\ x\text{ if 0}\le x\text{<1}\\ \text{2 if }x=\text{1}\end{array}\right.$

Calculation:

The left-hand limit at point $c$ exists when there is no graph to right of $x=c$ and when right hand limit is $\infty /-\infty$ .

Thus, the left-hand limit of $\underset{x\to c}{\lim }f\left(x\right)$ at point $c$ exists at $x=1$ .

d)

To find: The points $c$ where only the right-hand limit exists.

The right-hand limit of $\underset{x\to c}{\lim }f\left(x\right)$ at point $c$ exists at $x=-1$ .

Given information: The given piecewise defined function is $f\left(x\right)=\left\{\begin{array}{l}\sqrt{1-x^2}\text{ if -1}\le x<0\\ x\text{ if 0}\le x\text{<1}\\ \text{2 if }x=\text{1}\end{array}\right.$

Calculation:

The right-hand limit at point $c$ exists when there is no graph to left of $x=c$ and when left hand limit is $\infty /-\infty$ .

Thus, the right-hand limit of $\underset{x\to c}{\lim }f\left(x\right)$ at point $c$ exists at $x=1$ .