Question

Show that f(x) = x^1/3 is not differentiable at x = 0.

Answer 1

Disclaimer: It might be a wrong question because f(x) is differentiable at x=0 

Given: 

\[f(x) = x^\frac{1}{3}\]

We have,
(LHD at x = 0)

\[\lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{0 - h - 0}\]
\[ = \lim_{h \to 0} \frac{\left( 0 - h \right)^\frac{1}{3} - 0^\frac{1}{3}}{- h}\]
\[ = \lim_{h \to 0} \frac{\left( - h \right)^\frac{1}{3}}{- h}\]
\[ = \lim_{h \to 0} \left( - h \right)^\frac{- 2}{3} \]
\[ = 0\]

(RHD at x = 0)

\[\lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0}\]
\[ = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{0 + h - 0}\]
\[ = \lim_{h \to 0} \frac{\left( 0 + h \right)^\frac{1}{3} - 0^\frac{1}{3}}{- h}\]
\[ = \lim_{h \to 0} \frac{h^\frac{1}{3}}{h}\]
\[ = \lim_{h \to 0} h^\frac{- 2}{3} \]
\[ = 0\]

LHD at (x = 0)= RHD at (x = 0)

Hence,  

\[f(x) = x^\frac{1}{3}\]