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Math Help - LImit with and exponent variable

  1. #1
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    LImit with and exponent variable

    What is the limit of the sequence LImit with and exponent variable-codecogseqn.gifas n approaches infinity.
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  2. #2
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    Hint: a_n=\left(1+\frac{2}{n+1}\right)^{n+1}/\left(1+\frac{2}{n+1}\right).
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  3. #3
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    Ok i understand how you got that.. but I'm not sure where to start. I keep getting that the limit is 1 which I know it is not.
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  4. #4
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    Here is a useful equation:

    \displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x} = e^r.

    For your problem, instead of evaluating the limit of the ratio, evaluate the ratio of the limits:

    \displaystyle \lim_{n\to\infty} \dfrac{ \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1} }{1 + \dfrac{2}{n+1}} = \dfrac{ \displaystyle \lim_{n\to\infty} \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1} }{ \displaystyle \lim_{n\to\infty} \left ( {1 + \dfrac{2}{n+1}} \right ) }

    The limit in the denominator is 1. Division by 1 won't change anything, so the answer to your original problem is just the limit in the numerator.

    \displaystyle \lim_{n\to\infty} \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1}

    If we let r = 2 and x = n + 1, then the limit in your problem becomes \displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x}. Remember the useful equation at the top?

    The limit is just e^r. Substituting r = 2 gives: e^2.
    Last edited by NOX Andrew; February 26th 2011 at 06:22 PM.
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  5. #5
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    Thank you so much , it makes sense now.

    My mistake was not evaluating the numerator correctly. I found the limit of the denominator correctly, but for the numerator I said that the limit when n approaches infinity was = (1)^ infinity which is the same as 1. What was wrong with this reasoning?
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  6. #6
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    1^\infty is what is called an indeterminate form. It's value cannot be determined. In this problem, it turns out it is "equal" to e^2. In another problem, it might turn out to be "equal" to sin(1). Some other common indeterminate forms are \frac{0}{0} and \infty - \infty. Here is a link to a Wolfram article on indeterminate forms: Indeterminate -- from Wolfram MathWorld
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  7. #7
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    Oh , thats right. I completely forgot that it was indeterminate.
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