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Homework Help: A limit question

  1. Oct 20, 2007 #1
    i added a file with the question and how i tried to solve it

    lim [2+n^(1/3)]^(1/n)

    in the end i always get (infinity)^0 type of object
    then i got stuck
    what to do next??
     

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  3. Oct 20, 2007 #2
    I am unable to see your file, so what limit are you looking for?

    [tex]\lim_{x \to 0^-} [2+x^{\frac{1}{3}}]^{\frac{1}{x}} [/tex] or

    [tex]\lim_{x \to 0^+} [2+x^{\frac{1}{3}}]^{\frac{1}{x}} [/tex] or

    [tex]\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^{\frac{1}{x}} [/tex] ?

    Edit:
    Just a thought- let [tex]p=\lim_{x \to \infty} \frac{1}{x}[/tex] and evaluate. Then substitute into [tex]\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p [/tex] and evaluate.
     
    Last edited: Oct 20, 2007
  4. Oct 20, 2007 #3

    nrqed

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    As n goes to infinity, this goes to 1.

    As n gets very large, you may approximate this by [itex] n^{1/(3n)}[/itex]. taking the natural log of this (or log in any base), you can show that the ln goes to 0 as n goes to infinity. hence the initial function tends to e^0 = 1.
     
  5. Oct 20, 2007 #4
    ooohh sorry i forgot a part in the limit

    its

    lim [2+n^(1/3)]^(1/n)
    n>>infinity


    and i did got

    lim n^[1/(3n)]

    you say i need to take log

    it will be

    log X =1/(3n)
    n

    i didnt understand how to solve it that way
     
    Last edited: Oct 20, 2007
  6. Oct 20, 2007 #5
    Great.

    Then, as I mentioned earlier, evaluate:
    [tex]p=\lim_{x \to \infty} \frac{1}{x}[/tex]

    And plug into:
    [tex]\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p [/tex]

    Any non-zero base, raised to the power of [tex]0[/tex], is...
     
  7. Oct 20, 2007 #6
    your first limit P gives me 0

    the second one is infinity

    and infinity^0 is not 1
    its undefined
     
  8. Oct 20, 2007 #7

    nrqed

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    You need to consider the limit of [itex] \frac{\ln n}{n} [/itex]. The limit of this as n goes to infinity is zero. This is one of the basic limits and you probably can find the proof in yoru textbook.
     
  9. Oct 20, 2007 #8
    For the second limit atqamar posted [tex]\lim_{x \to \infty} [2+x^{\frac{1}{3}}]^p [/tex] is going to be 1 because anything other than zero raised to the 0 power is 1. How did you get undefined?

    Edit: How did you get infinity to the zero power? Thats not what the base is going to be
     
    Last edited: Oct 20, 2007
  10. Oct 20, 2007 #9
    the base is :

    2+ x^(1/3)

    if x gose to infinity the the base gose to infinity

    and p=0

    it gives me (infinity)^0

    and its undefined
     
  11. Oct 20, 2007 #10

    dynamicsolo

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    It is undefined because this is what is called an "indeterminate power"; other examples are 0^0 and 1^infinity. Nonetheless there will be a value for it.

    The solution involves one of the more exotic applications of L'Hopital's Rule, since this first has to be wrestled into the form of an indeterminate ratio 0/0 or infinity/infinity.

    You start off by taking the logarithm of your expression, which will be

    lim n-> inf. (1/n) ln [2 + n^(1/3)] ; this now has the apparent result 0 · infinity (an indeterminate product).

    You next rewrite this product as a ratio, since f · g = f / (1/g) ; it is usually best to put the logarithmic expression in the numerator to apply L'Hopital's Rule, since we're going to be taking derivatives:

    lim n-> inf. ( ln [2 + n^(1/3) ] ) / n .

    This now has the apparent value infinity/infinity, making it a candidate for L'Hopital's Rule. Differentiate numerator and denominator separately:

    lim n-> inf. [ ( 1/{ 2 + n^(1/3) } ) · ( 1/{ 3 · n^(2/3) } ) ] / 1 [using the Chain Rule on the numerator]

    Since n appears in the denominator raised to positive powers, once this compound fraction is simplified, the limit is (1/2)·(1/3)·0 = 0 .

    This, however, is the natural logarithm of the original expression, so this limit is the natural logarithm of our original limit. So the original limit is e^0 = 1 .

    This is a formal technique for dealing with such problems; as nrqed and atqamar point out, though, this one can sort of be "eyeballed" for a solution...
     
    Last edited: Oct 20, 2007
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