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Math Help - Tricky limit without using l'Hopital's rule

  1. #1
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    Tricky limit without using l'Hopital's rule

    It goes like this:
    lim[x->infinity](1 + (x + 1)/(4x^2 + 4x + 4))^(4x+1).

    The solution is e, that I got with Mathematica and by myself, using a few applications of l'Hopital's rule, but the calculation is very long and tedious...

    Is there any other way to solve it? Could we use the known equality lim[n->infinity](1+1/n)^n = e?
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  2. #2
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    e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1  }

    Is this what you're trying to verify? So basically you're trying to show this is equal to e by transforming it into \left(1+\frac{1}{u}\right)^u? You've been asked to do this without using logarithms or L-hopitals rule?
    Last edited by adkinsjr; December 14th 2010 at 03:37 PM.
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  3. #3
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    Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

    e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1  }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2  +4x+4}\right)

    But then we have again a case of 0*\infty, and another application of L'Hopital's rule is needed, and then another and... then, some derivatives need to be calculated and finally, when all is done, I get the right result e.

    I have a hunch there exists a more elegant path to this result, but unfortunately I don't see it at the moment. Care to help?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by soulforged View Post
    Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

    e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1  }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2  +4x+4}\right)

    But then we have again a case of 0*\infty, and another application of L'Hopital's rule is needed, and then another and... then, some derivatives need to be calculated and finally, when all is done, I get the right result e.

    I have a hunch there exists a more elegant path to this result, but unfortunately I don't see it at the moment. Care to help?
    Quick question: are you allowed to assume that \lim\limits_{x\to\infty}\left(1+\frac{1}{x}\right)  ^x=e? If so, then your limit can be computed in several steps.
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    Quote Originally Posted by soulforged View Post
    Well, I've tried using logarithms (a special case of L'Hopital's rule, actually), like this:

    e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1  }=\lim_{x->\infty}\left(1+4x)\log_{e}\left(1+\frac{1+x}{4x^2  +4x+4}\right)

    But then we have again a case of 0*\infty, and another application of L'Hopital's rule is needed, and then another and... then, some derivatives need to be calculated and finally, when all is done, I get the right result e.

    I have a hunch there exists a more elegant path to this result, but unfortunately I don't see it at the moment. Care to help?
    L'Hospital's Rule is often the most elegant path...
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    Quick question: are you allowed to assume that \lim\limits_{x\to\infty}\left(1+\frac{1}{x}\right)  ^x=e? If so, then your limit can be computed in several steps.
    Yep, I'm allowed to do that, and this is actually just what I'm looking for!

    Could you give a hint or two on how to solve it using this fact?
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  7. #7
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    Quote Originally Posted by adkinsjr View Post
    e=\lim_{x->\infty}\left(1+\frac{1+x}{4x^2+4x+4}\right)^{4x+1  }

    Is this what you're trying to verify? So basically you're trying to show this is equal to e by transforming it into \left(1+\frac{1}{u}\right)^u? You've been asked to do this without using logarithms or L-hopitals rule?
    \displaystyle\frac{x+1}{4x^2+4x+4}=\frac{1}{4x+\fr  ac{4}{x+1}}

    \displaystyle\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]

    As x\rightarrow\infty the 2nd factor \rightarrow\ 1

    and

    \displaystyle\ 4x+\frac{4}{x+1}\rightarrow\ 4x
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  8. #8
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    I guess this makes the problem solved. Thanks.

    Although I must admit I'm not entirely comfortable with this solution, because you've jumped from lim(f(x)^g(x)) to lim(f(x))^g(x), and I'm not sure if you can actually do that.

    In other words, in the last step, you've calculated the inner limit of (4x + 4/x+1) = 4x, and left the outer expression unafected, but we have an exponentiation there...

    I know one can do e.g. lim(f(x)*g(x)) = lim(f(x)) * lim(g(x)), but I know of no theorem that says something along similar lines for the exponentiation... if you see what I'm getting at.
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  9. #9
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    If we write for x>3

    \displaystyle\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}>\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+\frac{4}{x+1}}>\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}

    but

    \displaystyle\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x}

    therefore

    \displaystyle\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+1}=\lim_{x \to \infty}\left[1+\frac{1}{4x+\frac{4}{x+1}}\right]^{4x+\frac{4}{x+1}}
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  10. #10
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    Alright then, I get it now. Thank you very much for this explanation.
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