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    Find derivative of \vert x \vert and \vert x \vert^3 at x = 0 using the definition of the derivative.

    For \vert x \vert at x = 0 :

    After simplifying i got
    \displaystyle\lim_{h \to 0} \dfrac{\vert h \vert}{h} = \displaystyle\lim_{h \to 0} \dfrac{\sqrt{h^2}}{h}  = \displaystyle\lim_{h \to 0} 1 = 1}

    Since \vert h \vert = \sqrt{h^2}

    For \vert x \vert^3 at x = 0 :

    After simplifying i got
    \displaystyle\lim_{h \to 0} \dfrac{\vert h \vert^3}{h} =  \dfrac{(\sqrt{h^2})^3}{h} = \displaystyle\lim_{h \to 0}h^2 = 0

    Now i know that the derivative of \vert x \vert at x = 0 does not exist, but the derivative of derivative of \vert x \vert^3 at x = 0 is 0. Why doesn't what i have done work for both?
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    (Original post by John taylor)
    Find derivative of \vert x \vert and \vert x \vert^3 at x = 0 using the definition of the derivative.

    For \vert x \vert at x = 0 :

    After simplifying i got
    \displaystyle\lim_{h \to 0} \dfrac{\vert h \vert}{h} = \displaystyle\lim_{h \to 0} \dfrac{\sqrt{h^2}}{h}  = \displaystyle\lim_{h \to 0} 1 = 1}

    Since \vert h \vert = \sqrt{h^2}

    For \vert x \vert^3 at x = 0 :

    After simplifying i got
    \displaystyle\lim_{h \to 0} \dfrac{\vert h \vert^3}{h} =  \dfrac{(\sqrt{h^2})^3}{h} = \displaystyle\lim_{h \to 0}h^2 = 0

    Now i know that the derivative of \vert x \vert at x = 0 does not exist, but the derivative of derivative of \vert x \vert^3 at x = 0 is 0. Why doesn't what i have done work for both?
    Your mistake is in saying that \sqrt{h^2}=h - can you see what it should be?
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    (Original post by Mark13)
    Your mistake is in saying that \sqrt{h^2}=h - can you see what it should be?
    Why is it a mistake, isn't this true by definition and how come it works for the second one if its wrong?

    And i guess you mean \sqrt{h^2}= \vert h \vert
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    (Original post by John taylor)
    Why is it a mistake, isn't this true by definition and how come it works for the second one if its wrong?

    And i guess you mean \sqrt{h^2}= \vert h \vert
    The problem is, it isn't true. You're saying

    \dfrac{|h|}{h} = \dfrac{\sqrt{h^2}}{h} = 1

    But what if h = -1?
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    (Original post by Noble.)
    The problem is, it isn't true. You're saying

    \dfrac{|h|}{h} = \dfrac{\sqrt{h^2}}{h} = 1

    But what if h = -1?
    How come it works for the derivative of \vert x \vert^3
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    (Original post by John taylor)
    How come it works for the derivative of \vert x \vert^3
    Sometimes incorrect maths gives you the correct answer.
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    (Original post by John taylor)
    After simplifying i got
    \displaystyle\lim_{h \to 0} \dfrac{\vert h \vert}{h} = \displaystyle\lim_{h \to 0} \dfrac{\sqrt{h^2}}{h}  = \displaystyle\lim_{h \to 0} 1 = 1}
    This bit is wrong. Basically, the problem is apparent when you unpack the definition of what a limit really is. What fails is that there is no open set containing  0 where the h in your denominator is always positive (what you tacitly assume in making your erroneous conclusion.

    The easiest way to expand upon this is to write:

    |x|= \begin{cases}x, & x>0 \\ 0, & x=0 \\ -x, & x < 0\end{cases}

    and consider the one sided limits

    \mathrm{lim}_{x \rightarrow 0^-} \frac{|x| - |0|}{x - 0} = \mathrm{lim}_{x \rightarrow 0^-} \frac{-x}{x}=-1

    \mathrm{lim}_{x \rightarrow 0^+} \frac{|x| - |0|}{x - 0} = \mathrm{lim}_{x \rightarrow 0^-} \frac{x}{x}=1.

    Note that these differ, which means that the two sided limit

    \mathrm{lim}_{x \rightarrow 0} \frac{|x| - |0|}{x - 0}

    doesn't exist.
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    (Original post by Mark85)
    This bit is wrong. Basically, the problem is apparent when you unpack the definition of what a limit really is. What fails is that there is no open set containing  0 where the h in your denominator is always positive (what you tacitly assume in making your erroneous conclusion.

    The easiest way to expand upon this is to write:

    |x|= \begin{cases}x, & x>0 \\ 0, & x=0 \\ -x, & x < 0\end{cases}

    and consider the one sided limits

    \mathrm{lim}_{x \rightarrow 0^-} \frac{|x| - |0|}{x - 0} = \mathrm{lim}_{x \rightarrow 0^-} \frac{-x}{x}=-1

    \mathrm{lim}_{x \rightarrow 0^+} \frac{|x| - |0|}{x - 0} = \mathrm{lim}_{x \rightarrow 0^-} \frac{x}{x}=1.

    Note that these differ, which means that the two sided limit

    \mathrm{lim}_{x \rightarrow 0} \frac{|x| - |0|}{x - 0}

    doesn't exist.
    For the derivative of \vert x \vert would i need to still consider left and right limit of 0, even though it exists.
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    (Original post by John taylor)
    For the derivative of \vert x \vert would i need to still consider left and right limit of 0, even though it exists.
    I assume you mean |x|^3 and that is how you conclude the limit exists.

    f is differentiable at x_0 and f'(x_0) = l if and only if f has both left- and right-derivatives at x_0 and f'_-(x_0) = l = f'_+(x_0)
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    (Original post by Noble.)
    I assume you mean |x|^3 and that is how you conclude the limit exists.

    f is differentiable at x_0 and f'(x_0) = l if and only if f has both left- and right-derivatives at x_0 and f'_-(x_0) = l = f'_+(x_0)
    So for \vert x \vert^3 i got

     \displaystyle\lim_{h \to 0^-} -h^2 = 0

     \displaystyle\lim_{h \to 0^+} h^2 = 0

    so the the derivative exists and is 0 at x = 0, since both the left and right limit are the same. Is this right?
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    (Original post by John taylor)
    So for \vert x \vert^3 i got

     \displaystyle\lim_{h \to 0^-} -h^2 = 0

     \displaystyle\lim_{h \to 0^+} h^2 = 0

    so the the derivative exists and is 0 at x = 0, since both the left and right limit are the same. Is this right?
    Yes.
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    (Original post by John taylor)
    For the derivative of \vert x \vert would i need to still consider left and right limit of 0, even though it exists.
    Although what Noble writes above is true and answers the question you were asking I should add:

    One sided limits are a tool... you don't need them to define a general limit. What Noble posted is a useful theorem; not a definition. I mean, go back and look at the actual definition of a limit - no one sided limits are mentioned.

    The point is, for certain functions, it is useful to consider limiting behaviour when we think of our variable approaching some point, say, c seperately from the left and right. For example, if the function is defined piecewise

    e.g.

    f(x) = \begin{cases} f_1(x), & x < c \\ 0, & x = c \\ f_2(x), & x >c \end{cases}

    then we can write down simple formulae when dealing with the right and left one sided limits in terms of f_1 and f_2.

    In general, when considering the (two sided) limit at c, we have to consider open intervals containing c so we can't just pick f_1 or f_2 and base our calculations off that. That was essentially the mistake you made in your erroneous example.

    The take home message here is: go read, learn and inwardly digest the definition of a limit else you are wasting your time 'doing' such examples and asking questions about them.
 
 
 
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