Chapter 2, Problem 15RQ

To determine

To find: Whether the statement “If $f$ is continuous at $a$ , then $f$ is differential at $a$ ” is true or false.

Answer to Problem 15RQ

The statement “If $f$ is continuous at $a$ , then $f$ is differential at $a$ ” false.

Explanation of Solution

Given information:

The given statement is “ $f$ is continuous at $a$ ”.

Calculation:

Let us assume a function $f\left(x\right)=\left|x\right|$ .

Check the continuity of function for $x<0$ .

  $\begin{array}{l}\underset{x\to 0^{-}}{\text{lt}}f\left(x\right)=\underset{x\to 0^{-}}{\text{lt}}\left|x\right|\\ \underset{x\to 0^{-}}{\text{lt}}f\left(x\right)=\underset{x\to 0^{-}}{\text{lt}}\left(\begin{array}{l}-x\\ =\end{array}\right)\\ \underset{x\to 0^{-}}{\text{lt}}f\left(x\right)=\underset{h\to 0^{-}}{\text{lt}}-\left(0-h\right)\\ \underset{h\to 0^{-}}{\text{lt}}=0\end{array}$

Check the continuity of function for $x>0$ .

  $\begin{array}{l}\underset{x\to 0^{+}}{\text{lt}}f\left(x\right)=\underset{x\to 0^{+}}{\text{lt}}\left|x\right|\\ \underset{x\to 0^{+}}{\text{lt}}f\left(x\right)=\underset{x\to 0^{+}}{\text{lt}}\left(x\right)\\ \underset{x\to 0^{+}}{\text{lt}}f\left(x\right)=\underset{h\to 0^{-}}{\text{lt}}\left(0+h\right)\\ \underset{h\to 0^{+}}{\text{lt}}=0\end{array}$

Here, left hand limit is same as right hand limit and $f\left(0\right)=0$ .

So, the function is continuous at $0$ .

Evaluate left hand derivative of the function.

  $\begin{array}{c}L{f}^{\prime }\left(0\right)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{\left|x\right|-0}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{-x}{x}\\ =-1\end{array}$

Evaluate right hand derivative of the function.

  $\begin{array}{c}R{f}^{\prime }\left(0\right)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{\left|x\right|-0}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{x}{x}\\ =1\end{array}$

Since left hand derivative and right hand derivative is not same the given function is not differentiable.

Therefore, the statement “If $f$ is continuous at $a$ , then $f$ is differential at $a$ ” is false.