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

  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]\lim_{h\to_0}\frac{f\left(8+h\right)}{1}[/tex]

    then,

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

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

    matt grime 9,396
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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. i tried factoring according to this:

    [tex]a^4-b^4=\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)[/tex]

    but im stuck because i cant get a multiple of h to cancel the denomiator's h.
     
  5. matt grime

    matt grime 9,396
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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. ...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. dextercioby

    dextercioby 12,314
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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. Gib Z

    Gib Z 3,348
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    I think hes been asked to show that, at least for n= 4/3.
     
  9. 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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