Chapter 2.2, Problem 48E

a.

To determine

To show: The function $f$ is continuous.

Explanation of Solution

Given information:

The function is. $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$

Calculation:

The function is $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Continuity at $x=0$ .

Right hand limit of the function is as shown below.

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }x\sin \frac{1}{x}=0\times \sin \frac{1}{0}\\ =0\end{array}$

Left hand limit of the function is as shown below.

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }x\sin \frac{1}{x}=0\times \sin \frac{1}{0}\\ =0\end{array}$

Conclusion: Since, the right land limit , left hand limit and limit at a given point are equal. So, the function continuous at $x=0$ .

b.

To determine

To show: The value of $\frac{f\left(0+h\right)-f\left(0\right)}{h}=\sin \frac{1}{h}$

Explanation of Solution

Given information:

The function is. $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$

Calculation:

The function is $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Substitute the function in the left hand side,

  $\begin{array}{c}\frac{f\left(0+h\right)-f\left(0\right)}{h}=\frac{\left(0+h\right)\sin \left(\frac{1}{0+h}\right)-0\sin \left(\frac{1}{0}\right)}{h}\\ =\frac{h\sin \left(\frac{1}{h}\right)-0}{h}\\ =\frac{h\sin \left(\frac{1}{h}\right)}{h}\\ =\sin \left(\frac{1}{h}\right)\end{array}$

Conclusion: Thus, $\frac{f\left(0+h\right)-f\left(0\right)}{h}=\sin \frac{1}{h}$

c.

To determine

Explanation of Solution

Given information:

The function is. $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$

Calculation:

The function is $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Substitute the function in the left hand side,

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\frac{\left(0+h\right)\sin \left(\frac{1}{0+h}\right)-0\sin \left(\frac{1}{0}\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{h\sin \left(\frac{1}{h}\right)-0}{h}\\ =\underset{h\to 0}{\lim }\frac{h\sin \left(\frac{1}{h}\right)}{h}\\ =\underset{h\to 0}{\lim }\sin \left(\frac{1}{h}\right)\end{array}$

Solve further.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\sin \left(\frac{1}{h}\right)\\ =\sin \left(\frac{1}{0}\right)\\ =undefined\end{array}$

Conclusion: Thus, $\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ does not exist.

d.

To determine

To calculate: The left hand derivative and right hand derivative at $x=0$ .

Answer to Problem 48E

The right hand derivative and left hand derivative exist.

Explanation of Solution

Given information:

The function is. $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Concept used:

The left hand derivative is $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ and $\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$

Calculation:

The function is $f\left(x\right)=\left\{\begin{array}{c}x\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Right hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\left(0+h\right)\sin \frac{1}{0+h}-0\sin \frac{1}{0}}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{h\sin \frac{1}{h}-0}{h}\\ =\underset{h\to 0^{+}}{\lim }\sin \frac{1}{h}\\ =\infty \end{array}$

Left hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\left(0+h\right)\sin \frac{1}{0+h}-0\sin \frac{1}{0}}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h\sin \frac{1}{h}-0}{h}\\ =\underset{h\to 0^{-}}{\lim }\sin \frac{1}{h}\\ =\infty \end{array}$

Conclusion: So, both right and left derivates exist.

e.

To determine

The function is differentiable at $x=0$ .

Answer to Problem 48E

The derivative exists.

Explanation of Solution

Given information:

The function is $g\left(x\right)=\left\{\begin{array}{c}{x}^2\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Concept used:

The right hand derivative , left hand derivative should be equal.

Calculation:

The function is $g\left(x\right)=\left\{\begin{array}{c}{x}^2\sin \frac{1}{x},x\ne 0\\ 0,x=0\end{array}\right\}$ .

Right hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{{\left(0+h\right)}^2\sin \frac{1}{0+h}-0\sin \frac{1}{0}}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{h^2\sin \frac{1}{h}-0}{h}\\ =\underset{h\to 0^{+}}{\lim }h\sin \frac{1}{h}\\ =0\end{array}$

Left hand derivative is as shown below.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{{\left(0+h\right)}^2\sin \frac{1}{0+h}-0\sin \frac{1}{0}}{h}\\ =\underset{h\to 0^{-}}{\lim }\frac{h^2\sin \frac{1}{h}-0}{h}\\ =\underset{h\to 0^{-}}{\lim }h\sin \frac{1}{h}\\ =0\end{array}$

Also $g^{\prime }\left(0\right)=0$

Conclusion: As the left hand derivative, right hand derivative and $g^{\prime }\left(0\right)$ are equal to $0$ .