 David Young

2021-12-26

Q: Let $\left\{\begin{array}{cc}{x}^{m}\mathrm{sin}\frac{1}{x};& x\ne 0\\ 0;& x=0\end{array}$
Find the set of values of m for which
(i) $f\left(x\right)$ is continuous at $x=0$
(ii) $f\left(x\right)$ is differentiable at $x=0$
(iii) $f\left(x\right)$ is continuous but not differentiable at $x=0$. Bernard Lacey

The given function is:
$\left\{\begin{array}{cc}{x}^{m}\mathrm{sin}\frac{1}{x};& x\ne 0\\ 0;& x=0\end{array}$
i) f is continuous at $x=0$ if
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
Now, $\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}{x}^{m}\mathrm{sin}\frac{1}{x}$
$x=0$ is $m>0$
$\underset{x\to 0}{lim}f\left(0\right)\ne 0$ if $m<0$
${\delta }_{0}$ f is continuous for $mt\left(0,\mathrm{\infty }\right)$ Jenny Bolton

ii) Now, ${f}^{\prime }\left(x\right)=m\cdot {x}^{m-1}\cdot \mathrm{sin}\frac{1}{x}+{x}^{m}\cdot \mathrm{cos}\left(\frac{1}{x}\right)\cdot \left(\frac{-1}{{x}^{2}}\right)$
$m{x}^{m-1}\cdot \mathrm{sin}\left(\frac{1}{x}\right)-{x}^{m-2}\mathrm{cos}\left(\frac{1}{x}\right)$ if $x\ne 0$
So, it is differentiable at $x=0$
if $m>2$ because then only
f(x) is differentiable
Hence, f is continuous but only differentiable at $x=0$ if $m\in \left[2,\mathrm{\infty }\right)$ karton

iii) So, f is continuous but only differentiable at
$\left(0,\mathrm{\infty }\right);\left[2,\mathrm{\infty }\right)$
is on (0, 2)

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