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How do I prove that both are equivalent limits

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    If k is a positive integer, then show that
    ##\lim_{x\to\infty} (1+\frac{k}{x})^x = \lim_{x\to 0} (1+kx)^\frac{1}{x}##

    2. Relevant equations
    L'Hopitals rule, Taylor's expansion

    3. The attempt at a solution
    How should I begin? Should I prove that both has the same limit, or is there another way to work this question out?
     
  2. jcsd
  3. Jan 30, 2016 #2

    QuantumQuest

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    Gold Member

    You can do the Taylor expansions to some number of terms for both limits and see - easily enough, that they converge to the same limit.
     
  4. Jan 30, 2016 #3

    Mark44

    Staff: Mentor

    A simpler approach than the one suggested by QuantumQuest is use a simple substitution.
     
  5. Jan 30, 2016 #4
    What substitution do you mean?
     
  6. Jan 30, 2016 #5

    blue_leaf77

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    Science Advisor
    Homework Helper

    When you compare
    $$(1+\frac{k}{x})^x$$
    and
    $$(1+kx)^\frac{1}{x},$$
    how can you go from one to the other?
     
  7. Jan 30, 2016 #6

    Mark44

    Staff: Mentor

    Maybe think about them like this:
    ##(1+\frac{k}{x})^x##
    and
    ##(1+ku)^\frac{1}{u}##
     
  8. Jan 30, 2016 #7
    Will this work as a proof?

    ##L_1=\lim_{x\to\infty} (1+\frac{k}{x})^x##
    ##\ln L_1=\ln[\lim_{x\to\infty} (1+\frac{1}{k})^x]##
    ##\ln L_1=\lim_{x\to\infty} \ln[(1+\frac{1}{k})^x]##


    ##\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-...##
    ##\ln(1+\frac{k}{x})=\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...##

    ##=\lim_{x\to\infty} x [\frac{k}{x}-\frac{k^2}{2x^2}+\frac{k^3}{3x^3}-...]##
    ##=\lim_{x\to\infty} [k-\frac{k}{2x^2}+\frac{k^2}{3x^3}-...]##
    ##=k##

    Since
    ##\ln L_1=k##

    Therefore
    ##L_1=e^k##

    For the right hand side
    ##L_2=\lim_{x\to 0} (1+kx)^\frac{1}{x}##
    ##\ln L_2=\ln\lim_{x\to 0} (1+kx)^\frac{1}{x}##
    ##\ln L_2=\lim_{x\to 0} \ln (1+kx)^\frac{1}{x}##
    ##\ln L_2=\lim_{x\to 0} \frac{1}{x} \ln (1+kx)##
    ##\ln L_2=\lim_{x\to 0} \frac{1}{x} (kx-\frac{k^2x^2}{2}+\frac{k^3 x^3}{3}-...)##
    ##\ln L_2=\lim_{x\to 0} (k-\frac{k^2x}{2}+\frac{k^3 x^2}{3}-...)##
    ##\ln L_2=k##
    ##L_2=e^k##

    Since both converges to the same limit, therefore both limits are equivalent.
     
    Last edited: Jan 30, 2016
  9. Jan 31, 2016 #8

    Mark44

    Staff: Mentor

    That's an awful lot of work to prove something that's very straightforward. If ##x \to \infty## then ##\frac 1 x \to 0##. Does that give you an idea of the substitution to do?
     
  10. Jan 31, 2016 #9

    blue_leaf77

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    Science Advisor
    Homework Helper

    You can also do it that way, although as Mark hinted above, substitution may lead to significant reduction in the steps needed to do the proof.
     
  11. Jan 31, 2016 #10
    u=1/x?
     
  12. Jan 31, 2016 #11
    I think I got the answer - do check if I commit any mistakes:

    ##x=\frac{1}{u}##, ##u=\frac{1}{x}##
    When ##u \rightarrow \infty ##, ##x \rightarrow 0##

    Then the limit can be rewritten as
    ##\lim_{u\to 0} (1+ku)^\frac{1}{u}##

    Replacing u with x gives
    ##\lim_{x\to 0} (1+kx)^\frac{1}{x}##

    Which is equal to the right hand side
     
  13. Jan 31, 2016 #12

    blue_leaf77

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    Science Advisor
    Homework Helper

    Yes that's right.
     
  14. Jan 31, 2016 #13
    Great! Can anyone mark this thread as solved? ;)
     
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