Question

The function $f\left(x\right)$ is given by $f\left(x\right)=(\begin{array}{c}{x}^2\sin \left(\frac{1}{x}\right),\mathrm{if}x\ne 0\\ 0,\mathrm{if}x=0\end{array}$
Which of the following statement(s) is/are true for $f\left(x\right)?$

• A. $f\left(x\right)$ is continuous at x = 0
• B. $f\left(x\right)$ is differentiable at x = 0
• C. $f^{\prime }\left(0\right)=0$
• D. $f\left(x\right)$ is not differentiable at x = 0

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)$ is continuous at x = 0
B $f\left(x\right)$ is differentiable at x = 0
C $f^{\prime }\left(0\right)=0$

Given that
$f\left(x\right)=(\begin{array}{c}{x}^2\sin \left(\frac{1}{x}\right),\mathrm{if}x\ne 0\\ 0,\mathrm{if}x=0\end{array}\underset{x\to 0}{\lim }{x}^2=0\mathrm{and}-1\le \sin (\frac{1}{x})\le 1-x^2\le x^2\sin (\frac{1}{x})\le x^2\underset{x\to 0}{\lim }-x^2\le \underset{x\to 0}{\lim }{x}^2\sin (\frac{1}{x})\le \underset{x\to 0}{\lim }{x}^2\mathrm{By}\mathrm{Sandwitch}\mathrm{Theorem},\underset{x\to 0}{\lim }{x}^2\sin (\frac{1}{x})=0\therefore f(x)\mathrm{is}\mathrm{continuous}\mathrm{at}x=0.\mathrm{Now},(\mathrm{RHD}\mathrm{at}x=0)=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{0+h-0}=\underset{h\to 0^{+}}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{{\left(h\right)}^2\sin \left(\frac{1}{h}\right)-0}{h}=\underset{h\to 0^{+}}{\lim }h\sin (\frac{1}{h})=0\times (\mathrm{an}\mathrm{oscillating}\mathrm{number}\mathrm{between}-1\mathrm{and}1)\Rightarrow (\mathrm{RHD}\mathrm{at}x=0)=0(\mathrm{LHD}\mathrm{at}x=0)=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{h\to 0^{-}}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0}=\underset{h\to 0^{-}}{\lim }\frac{f\left(-h\right)-f\left(0\right)}{-h}=\underset{h\to 0^{-}}{\lim }\frac{{\left(-h\right)}^2\sin \left(\frac{1}{-h}\right)-0}{-h}=\underset{h\to 0^{-}}{\lim }h\sin (\frac{1}{h})=0\times (\mathrm{an}\mathrm{oscillating}\mathrm{number}\mathrm{between}-1\mathrm{and}1)\Rightarrow (\mathrm{LHD}\mathrm{at}x=0)=0\therefore (\mathrm{LHD}\mathrm{at}x=0)=(\mathrm{RHD}\mathrm{at}x=0)\mathrm{So},f(x)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=0\mathrm{and}f\text{'}(0)=0$