My Math Forum Limit rules trick question?

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 December 12th, 2015, 02:19 PM #1 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 Limit rules trick question? Use the definition of the derivative (and limit rules) to find the derivative of |x^3| at x = 0 Is this a trick question? Like just from looking at it, I know the derivative of an absolute value such as x^3 does not equate at x=0, right?
 December 12th, 2015, 02:58 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $\displaystyle f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ $\displaystyle f'(0) = \lim_{x \to 0} \frac{|x^3| - 0}{x - 0}$ $x < 0 \implies |x^3| = -x^3$ $\displaystyle \lim_{x \to 0^-} \frac{-x^3}{x}$ $\displaystyle \lim_{x \to 0^-} -x^2 = 0$ $x \ge 0 \implies |x^3| = x^3$ $\displaystyle \lim_{x \to 0^+} \frac{x^3}{x}$ $\displaystyle \lim_{x \to 0^+} x^2 = 0$ limit from the left = limit from the right, therefore ... $\displaystyle f'(0) = \lim_{x \to 0} \frac{|x^3| - 0}{x - 0} = 0$
 December 12th, 2015, 04:53 PM #3 Member   Joined: May 2015 From: Earth Posts: 64 Thanks: 0 So you're saying apply the sandwich rule or whatever? To take the derivative on both sides to get the limit? Upon looking at the graph I see there is no edge there like I thought there was. If there was an edge, there would be no limit there, or derivative? How does that all I work? I forget. Thanks.
 December 12th, 2015, 06:23 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 It's straightforward to show from first principles that ax³ has derivative 3ax², with value 3a*0 = 0 at x = 0. Letting a = sgn(x) doesn't affect this result. That wouldn't work for |x|, as the derivatives of x and -x take different values at x = 0. The derivative of |x³| is 3x|x|, as can be verified by differentiating (x²)$^{3/2}$ using the usual rules.

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