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Please help me the this limit

  1. Mar 23, 2005 #1

    I stumble on this problem that I have no clue how to do. Can someone please help me. I was thinking that this problem might be somehow can be solve by taking the natural log and then e back again. But I am stuck somehow. :confused:
  2. jcsd
  3. Mar 23, 2005 #2


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    Make the substitution

    [tex] \frac{2}{x}=\frac{1}{u} [/tex]

  4. Mar 23, 2005 #3


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    lim(1+2/x)^x when x -> infinity
    At first: 1 + 2/x = (x + 2)/x
    You set:
    y = lim(1+2/x)^x when x -> infinity
    -> lny = ln (lim(1+2/x)^x when x -> infinity)
    Since ln(lim(1+2/x)^x when x -> infinity) is continuous function
    -> lny = lim[ln(1+2/x)^x when x -> infinity]
    -> lny = lim[xln(1+2/x) when x -> infinity]
    -> lny = lim[ln(1+2/x)/(1/x)]
    We can see that this limit has form: infinity/infinity
    -> We use L'Hopital rule, take derivative:
    derivative of ln(1+2/x) = (-2/x^2)/(1+2/x) = -2/[x(x+2)]
    derivative of 1/x = -1/x^2
    -> lny = lim [2x/(x+2)] when x -> infinity
    One more time use L'Hopital rule
    -> lny = lim (2/1) = 2
    -> y = e^2
    e^2 = lim(1+2/x)^x when x -> infinity
  5. Mar 23, 2005 #4
    thank you so much, you have just saved my life. Thank you for the detail explaination =)
  6. Mar 23, 2005 #5


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    [tex] x=2u [/tex]

    [tex] \lim_{x\rightarrow +\infty} \left(1+\frac{2}{x}\right)^{x}=\lim_{u\rightarrow +\infty} \left(1+\frac{1}{u}\right)^{2u}=e^{2} [/tex]

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