Q. 78. - 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. 

$f\left(x\right)=\left\{\begin{array}{l}{x}^2 \sin \frac{1}{x} , if x\ne 0\\ 0 , if x=0,\end{array}\right.x=0$.

Step-by-Step Solution

Answer

The function is continuous and differentiable at $x=0$.

Step 1. Given Information.

The given function is 

$f\left(x\right)=\left\{\begin{array}{l}{x}^2 \sin \frac{1}{x} , if x\ne 0\\ 0 , if x=0,\end{array}\right.x=0$.

Step 2. Calculation.

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

The Right hand function will be 

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

Step 3. Calculation.

The Left hand limit will be :

$\begin{array}{rcl}f\left(0^{-}\right)& =& f\left(0-h\right)\\ & =& h^2 \sin \left(\frac{1}{0-h}\right)\\ & =& \underset{h\to 0}{\lim }\frac{h^2\sin \left({\displaystyle \frac{1}{-h}}\right)}{-h}\\ & =& \underset{h\to 0}{\lim }h·\sin \left(\frac{1}{h}\right)\\ & =& 0·\left(-1 to 1\right)\\ & =& 0.\end{array}$

The function is continuous and differentiable at $x=0$,