Q. 77. - Calculus

Question

In Exercises 69–80, determine whether or not f is continuous and/or differentiable at the given value of x. If not, determine any left or right continuity or differentiability. For the last four functions, use graphs instead of the definition of the derivative.

$f\left(x\right)=\left\{\begin{array}{l}x\sin \frac{1}{x} , if x\ne 0\\ 0 , if x=0,\end{array}\right.x=0$

Step-by-Step Solution

Answer

The function $f\left(x\right)$ is continuous but not differentiable and function doesn't exists.

Step 1. Given Information.

The given function is 

$f\left(x\right)=\left\{\begin{array}{l}x\sin \frac{1}{x} , if x\ne 0\\ 0 , if x=0,\end{array}\right.x=0$.

Step 2. Differentiability.

The fucntion will be $f\left(0\right)=0$.

The Right hand function will be 

$\begin{array}{rcl}f\left(0^{+}\right)& =& f\left(0+h\right)\\ & =& \sin \left(\frac{1}{0+h}\right)\\ & =& \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{1}{x}}\right)}{\left(\frac{1}{x}\right)}\\ & =& 1\end{array}$

Step 3. Calculation.

The right hand function will be:

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

$\begin{array}{rcl}Rf\text{'}\left(0\right)& =& \underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{1}{0+h}}-0\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{h·\sin {\displaystyle \frac{1}{h}}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{1}{h}}}{1}, does not exists.\end{array}$

$f\left(x\right)$ is continuous but not differentiable function.$\begin{array}{rcl}Rf\text{'}\left(0\right)& =& \underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{1}{0+h}}\right)-0}{h}\\ & =& \underset{h\to 0}{\lim }\frac{h\sin {\displaystyle \frac{1}{h}}}{h}\\ & =& \underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{1}{h}}}{1}, does not exists.\\ & & \end{array}$