Show that f(x)=x^1 3 is not differentible at x = 0. - Vidyadip

Question

Show that $\text{f}(\text{x})=\text{x}^\frac{1}{3}$ is not differentible at x = 0.

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Answer

$\text{f}(\text{x})=\text{x}^\frac{1}{3}$
(LHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0-\text{h})-\text{f}(0)}{0-\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0^{-}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-\text{h})^\frac{1}{3}}{-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(-1)^\frac{1}{3}\text{h}^\frac{1}{3}}{(-1)-\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}(-1)^\frac{-2}{3}\text{h}^\frac{-2}{3}$
= Not defined
(RHL at x = 0) $=\lim_\limits{\text{x}\rightarrow0^{+}}\frac{\text{f}(\text{x})-\text{f}(0)}{\text{x}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{\text{f}(0+\text{h})-\text{f}(0)}{0+\text{h}-0}$
$=\lim_\limits{\text{x}\rightarrow0}\frac{(\text{h})^\frac{1}{3}-0}{\text{h}}$
$=\lim_\limits{\text{x}\rightarrow0}\text{h}^\frac{-2}{3}$
= Not defined
Since,
LHL and RHL does not exists at x = 0
$\therefore$ f(x) is not differentiable at x = 0

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