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Math Help - limits

  1. #1
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    limits

    Hi, okay so I have the following limit:

    lim x-->infinity [1+(4/(x-2))]^x

    so I don;'t know how to get rid of the power of x...? I am pretty sure that I have to put it into a l'hopitalable form but because of the power of x I am stuck!



    Thank you
    note: I triedwritting all this nicely with html codes but it said there aas an error sorry about that!
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  2. #2
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    Hi mate

    1 = (x-2)/(x-2)

    => 1+(4/(x-2)) = (x-2)/(x-2) + 4/(x-2) = (x+2)/(x-2)

    (x+2)/(x-2) = e^{ln [(x+2)/(x-2)]}


    [1+(4/(x-2))]^x = e^{x*ln [(x+2)/(x-2)]}

    Now you should be able to get the solution e^4, for x -> infty

    Yours,
    Rapha
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  3. #3
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    okay sorry I am still a little confused.. why did you equate 1 to (X-2)/ (x-2) ... I mean are we not supposed to l'hopital it at some point?

    Your answer is right so what you did is 100% correct I just don't understand it!
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  4. #4
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    Quote Originally Posted by thehabsonice View Post
    okay sorry I am still a little confused..
    why did you equate 1 to (X-2)/ (x-2) ...

    I wanted to get one term for 1+(4/(x-2)), that is why i equated to (x-2)/(x-2)

    Quote Originally Posted by thehabsonice View Post
    I mean are we not supposed to l'hopital it at some point?
    yes, we are, unless there is another solution, but I don't know an alternative.
    So

    my last step was



    I'm gonna show that x*ln [(x+2)/(x-2)] = 4, x to infty

    x*ln [(x+2)/(x-2)]  = \frac{1}{(x)^{-1}}*ln((x+2)/(x-2))

    Derivatives:

    (\frac{1}{(x)^{-1}})' = -1/x^2

    [ ln((x+2)/(x-2)) ] ' = 4/((x + 2)(2 - x))

    Using L'Hospital leeds to the problem

    - \frac{4x^2}{(x + 2)(2 - x)}


    Are you able to show that - \frac{4x^2}{(x + 2)(2 - x)} = 4, x to infty

    ?

    Quote Originally Posted by thehabsonice View Post
    Your answer is right so what you did is 100% correct I just don't understand it!
    Thank you, this statement is very polite


    Rapha
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  5. #5
    Moo
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    Hello,

    Otherwise, substitute t=x-2 \Rightarrow x=t+2

    So the limit becomes :

    \lim_{t\to\infty} \left(1+\frac 4t\right)^{t+2}=\lim_{t\to\infty} \left(1+\frac 4t\right)^t \cdot \left(1+\frac 4t\right)^2

    The first one is known : \lim_{x\to\infty}\left(1+\frac ax\right)^x=e^a
    and the second one just goes to 1.
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