Chapter 3.2, Problem 48E

(a)

To determine

Todetermine:Whether the function $f$ is continuous at $x=0$ .

Answer to Problem 48E

The function $f\left(x\right)$ is continuous at $x=0$ .

Explanation of Solution

Given information:

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

Formula Used:

Sandwich theorem:

If $f\left(x\right)\le g\left(x\right)\le h\left(x\right)$ and for $x=k$ if $f\left(x\right)=h\left(x\right)$ then $f\left(x\right)=g\left(x\right)=h\left(x\right)$ .

Calculation:

Consider $f$ is continuous a $x=0$ , then its limit at $x=0$ must be equal to $f\left(0\right)$ .

Thus, $\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)\\ =0\end{array}$

Use the sandwich theorem.

For any $x\ne 0$ , the inequality $x$ :

  $-x\le \sin \frac{1}{x}\le x$

Since, $\begin{array}{c}\underset{x\to 0}{\lim }x=\underset{x\to 0}{\lim }\left(-x\right)\\ =0\end{array}$

Thus, by the sandwich theorem.

  $\underset{x\to 0}{\lim }x\sin \frac{1}{x}=0$

Hence, the function $f\left(x\right)$ is continuous at $x=0$ .

(b)

To determine

To prove: $\frac{f\left(0+h\right)-f\left(0\right)}{h}=\sin \frac{1}{h}$ .

Answer to Problem 48E

Explanation of Solution

Given information:

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

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant

Proof:

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

The given expression is $\frac{f\left(0+h\right)-f\left(0\right)}{h}$

Substitute the value in the above expression.

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

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

Hence proved.

(c)

To determine

The reason why $\underset{x\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ .

Answer to Problem 48E

The limit $\underset{x\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ does not exist.

Explanation of Solution

Given information:

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

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

The value of $\sin \left(\frac{1}{h}\right)$ oscillates between $-1$ and 1.

Thus, at $h=0$ the function oscillates infinite number of times through the point 0

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

(d)

To determine

Whether the function $f$ has either a left hand or right hand derivative at $x=0$ .

Answer to Problem 48E

The value of $\underset{x\to 1^{+}}{\lim }2x$ is 2.

Explanation of Solution

Given information:

The given function: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\le 1\\ 2x,x>1\end{array}$

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

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

For the left hand derivative $x=0^{-}$ . Thus, the function will be $f\left(x\right)=x\sin \frac{1}{x}$ .

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

Similarly for the right hand derivative $x=0^{+}$ .

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

But from 48c, $\underset{x\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ does not exist at $x=0$ .

Hence, $f$ has neither a left hand nor a right hand derivative at $x=0$ .

(e)

To determine

To calculate:The function $g\left(x\right)$ is differentiable at $x=0$ and that $g\text{'}\left(0\right)=0$ .

Answer to Problem 48E

The function $g\left(x\right)$ is differentiable at $x=0$ and that $g\text{'}\left(0\right)=0$ .

Explanation of Solution

Given information:

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

Formula Used:

Definition of the derivative:

  $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Where $f\left(x\right)$ is the function and $h$ is a constant.

Calculation:

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

Find the derivative of $g\left(x\right)$ . Use the formula $f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

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

At $x=0$ , $g\left(x\right)$ is differentiable.

Hence, the function $g\left(x\right)$ is differentiable at $x=0$ and that $g\text{'}\left(0\right)=0$ .