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Can I use L'Hospitals rule here? (seem like i use it too often )

  1. Mar 13, 2007 #1
    Does l'hospitals rule work here?:

    [tex]\lim_{h\to_0}\frac{f\left(8+h\right)-f\left(8\right)}{h}[/tex] for [tex]f\left(x\right)=x^\frac{4}{3}[/tex]

    then i would get,



    [tex]=\frac{\left(8\right)^\frac{4}{3}}{1}[/tex] ?

    is that it?...
    Last edited: Mar 13, 2007
  2. jcsd
  3. Mar 13, 2007 #2

    matt grime

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    No, you can't use it there, in my opinion. The question is asking you to work out the derivative of x^{4/3} at x=8, thus you can't invoke l'Hopital which a priori assumes the function to be differentiable. A minor quibble, but given the way you've written it it seems clear that you're being asked to prove the derivative exists, so assuming it does is not allowed.
  4. Mar 13, 2007 #3
    i tried factoring according to this:


    but im stuck because i cant get a multiple of h to cancel the denomiator's h.
  5. Mar 13, 2007 #4

    matt grime

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    Where did the power of 1/3 vanish to? (Note I don't have a simple solution in mind. But it is easy to show that x^4 is differentiable, as is x^3, and hence x^1/3 by the inverse function theorem, hence x^{4/3} is diffible, and it all follows some what simply from these big sounding theorems).
  6. Mar 13, 2007 #5
    ...we haven't touched those theorems yet.

    but, [tex]a=\left(8+h\right)^\frac{1}{3}[/tex] and [tex]b=\left(8\right)^\frac{1}{3}[/tex]
  7. Mar 14, 2007 #6


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    x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n-1). Just plug n=4/3 and x=8 to see what you get.
  8. Mar 14, 2007 #7

    Gib Z

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    I think hes been asked to show that, at least for n= 4/3.
  9. Mar 14, 2007 #8
    I see it equals what i wrote using l'hospitals rule, but through what derivative what did you go through to determine that?
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