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Thread: limits

  1. #1
    Senior Member
    Apr 2009


    Consider the function

    1) Give a reason why the quotient rule for differentiation cannot be used to find g'(0)
    2) Determine whether or not g is differentiable at x=0 by evaluating separately the left and right hand limits:
    3) Why is it that derivatives do not exist at end points?
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  2. #2
    MHF Contributor
    Mar 2007


    1) Does the absolute-value part have a derivative everywhere? (Think about where the graph of y = |x| has an "elbow"....)

    2) For this part, just find the limits. Since |x| = -x for x < 0 and |x| = +x for x > 0, the limits will not have the same sign.

    3) Can you have two one-sided limits in agreement if you have only one one-sided limit?

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  3. #3
    Senior Member pankaj's Avatar
    Jul 2008
    New Delhi(India)
    It appears you got the definition of the Left hand derivative wrong.It goes like this

    2) g_{+}'(0)=\lim_{h\to 0}\frac{g(0+h)-g(0)}{h}=\lim_{h\to 0}\frac{g(h)}{h}=\lim_{h\to 0}\frac{1}{1-|h|}=1

    g_{-}'(0)=\lim_{h\to 0}\frac{g(0-h)-g(0)}{-h}=\lim_{h\to 0}\frac{g(-h)}{-h}=\lim_{h\to 0}\frac{1}{1-|-h|}=1

    So you see the derivative indeed exists.

    Using quotient rule we may get
     <br />
g'(x)=\frac{(1-|x|).1-x(-\frac{|x|}{x})}{(1-|x|)^2}=\frac{1-|x|+|x|}{(1-|x|)^2}=\frac{1}{(1-|x|)^2}<br />

    So you still get the derivative at x=0 as 1 but there is a slight hitch since
    would not exist at x=0,though by some kind of default we do get our answer.

    Let us see what other members have to say
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