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Question
Show that f(x) = x1/3 is not differentiable at x = 0.
Solution
Disclaimer: It might be a wrong question because f(x) is differentiable at x=0
Given:
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,
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