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Trying to figure out a limit as x-->a

  1. Jan 1, 2017 #1
    1. The problem statement, all variables and given/known data

    the limit x->a of [(x+2)^5/3 - (a+2)^5/3] / (x-a)

    2. Relevant equations

    limit laws

    3. The attempt at a solution

  2. jcsd
  3. Jan 1, 2017 #2


    Staff: Mentor

    Does the formula in the third attempt remind you on something?
  4. Jan 2, 2017 #3
    it looks like i'm trying to get a derivative. I am trying to do this without working using any derivatives and for some reason the power of 5/3 is completely messing with my head and i do not know how to proceed. i am still getting a 0/0 situation.
  5. Jan 2, 2017 #4


    Staff: Mentor

    Well, differentiation seems to be the shortest way. Otherwise you probably will have to follow the paths the differentiating formulas are proven. The power ##\frac{1}{3}## is the difficulty here, for you cannot expand it easily. Why do you want to restrict yourself, once you already have the formula for the first derivative?
  6. Jan 2, 2017 #5
    The restriction is due to the question is apparently brought up before any derivatives are introduced. So apparently this is do-able without using straight up differentiation. Yes, the 1/3 is killing me. and i have no idea how to approach it.
  7. Jan 2, 2017 #6


    Staff: Mentor

    I've just looked up how ##\frac{d}{dx} x^{\alpha}=\alpha x^{\alpha-1}## is proven for non-integer values ##\alpha##.
    It's done by the chain rule and ##x^\alpha = \exp(\alpha \ln x)##. Perhaps this might help and you may use properties of the exponential function. Another substitution ##y=x+2## should decrease writing work.
  8. Jan 2, 2017 #7
    Set ##X=(x+2)^{1/3}## ##A=(a+2)^{1/3}## then the it becomes ##\lim_{X\rightarrow A}{\frac{X^5-A^5}{X^3-A^3}}## . Hope I am right here and this helps.
  9. Jan 2, 2017 #8
    this seems to take care of it - provided you know how to expand a sum of powers to a high level - had to look it up to be honest:

    so revised sheet:
  10. Jan 7, 2017 #9


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    Correct result.
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