Show that the function f(x) x^m (1 x ), &x 0 0 ,& x=0 Continuous but not diffierentiable at x = 0, if 0 < m < 1

LHL= _ x 0^ - f(x) = _ h 0 f(0-h) = _ h 0 (-h)^m (-1 h ) = _ h 0 (-h)^m (1 h ) =0 [Where-1 1] =0 RHL = _ x 0^ + f(x) = _ h 0^+ f(0+h) = _ h 0 (+h)^m (1 0+h ) = _ h 0^- (-h)^m (1 h ) =0 ' [When-1 ' 1] =0 LHL = f(0) = RHL f(x) is continuous at x = 0 For differentiable at x = 0 (LHL at x = 0) = _ x 0^ - f(x)-f(0) x-0 = _ h 0 (0-h)-f(0) (0-h)-0 = _ h 0 (-h)^m (-1 h ) -h = _ h 0 (-h)^m-1 (-1 h ) = Not definded [Since 0 < m < 1] (RHL at x = 0) = _ x 0^ + f(x)-f(0) x-0 = _ h 0 f(0+h)-f(0) (0+h)-0 = _ h 0 h^m (1 h ) h = _ h 0 (h)^m-1 (1 h ) = Not defined [as 0, m < 1] (LHL at x = 0) and (RHL at x = 0) are not defined, so f(x) is continuous but not differentiable at x = 0, when 0 < m < 1.

Show that the function f(x) x^m (1 x ), &x 0 0 ,& x=0 Continuous …

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Show that the function $\text{f(x)}\begin{cases}\text{x}^\text{m}\sin(\frac{1}{\text{x}}), &\text{x}\neq0 \\0 ,& \text{x}=0\end{cases}$
Continuous but not diffierentiable at x = 0, if 0 < m < 1

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Answer

$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k}\ [\text{Where}-1\leq\text{k}\leq1]$
$=0$
$\text{RHL }=\lim_\limits{\text{x}\rightarrow0^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0^+}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(+\text{h})^\text{m}\sin\Big(\frac{1}{0+\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0^-}(-\text{h})^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)$
$=0\times\text{k'}\ [\text{When}-1\leq\text{k}'\leq1]$
$=0$
LHL = f(0) = RHL
$\therefore$ f(x) is continuous at x = 0
For differentiable at x = 0
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(0-\text{h})-\text{f}(0)}{(0-\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(-\text{h})^\text{m}\sin\Big(-\frac{1}{\text{h}}\Big)}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(-\text{h})^\text{m-1}\sin\Big(-\frac{1}{\text{h}}\Big)$
= Not definded [Since 0 < m < 1]
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{(0+\text{h})-0}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}^\text{m}\sin\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}(\text{h})^\text{m-1}\sin\Big(\frac{1}{\text{h}}\Big)$
= Not defined [as 0, m < 1]
$\therefore$ (LHL at x = 0) and (RHL at x = 0) are not defined, so f(x) is continuous but not differentiable at x = 0, when 0 < m < 1.

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