Question

Solve this:

$\mathrm{Q}.3\mathrm{A}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{is}\mathrm{defined}\mathrm{as}\mathrm{f}\left(\mathrm{x}\right)=[\begin{array}{cc}{\mathrm{x}}^{\mathrm{m}}\sin \frac{1}{2}\mathrm{x}\ne 0,& \mathrm{m}\in \mathrm{N}\\ 0\mathrm{If}& \mathrm{x}=0\end{array}.\mathrm{The}\mathrm{least}\mathrm{value}\mathrm{of}\mathrm{m}\mathrm{for}\mathrm{which}\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{is}\mathrm{continous}\mathrm{at}\mathrm{x}=0\mathrm{is}.$
(A) 1 (B) 2 (C) 3 (D) None.

Answer 1

$$\begin{gathered} forf\left(x\right)tobecontinuousatx=0 \\ \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)=0 \\ =\underset{x\to 0}{\lim }{x}^m\sin \frac{1}{x} \\ Note:\underset{x\to 0}{\lim }\sin \frac{1}{x}doesnotexist,wecannottellwhichvalueitwilltake.However,weknowthatitwillalwaysliebetween-1and1as\sin ealwaysliebetween-1and1 \\ \underset{x\to 0}{\lim }\sin \frac{1}{x}doesnotexistbutitisfinite.Soif\underset{x\to 0}{\lim }{x}^{m}existandisequaltozerothen.Somethingfinitemutlipliedbyzerowillgivezeroandthislimitwillhenceexist \\ For\underset{x\to 0}{\lim }{x}^m=0thepowerofxmustbepositive \\ so \\ m>0 \\ Ifm\in \mathrm{\mathbb{N}} \\ Thenleastvalueofm=1 \end{gathered}$$