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Math Help - Limit

  1. #1
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    Limit

    any tips on how to do this:

    lim(n->infinity) (2^n-5^n)/(3^n+5^n)

    thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello

    You can try to factor both the numerator and the denominator by 5^n.
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  3. #3
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    Thanks! got it
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by ah-bee View Post
    any tips on how to do this:

    lim(n->infinity) (2^n-5^n)/(3^n+5^n)

    thanks
    Always try to compare the "top" and "bottom" of a fraction. If the "bottom" is always more than the "top" and either the "top" is decreasing and/or the "bottom" is increasing, then the limit of such a fraction is zero.

    "Top" = 2^n - 5^n
    "Bottom" = 3^n + 5^n

    Let's see if:

    "Top" < "Bottom"
    2^n - 5^n < 3^n + 5^n

    2^n -3^n< 2*5^n

    Note that 3^n > 2^n, so the left side yields a negative result. This inequality is always true.

    Therefore, we can safely say that the limit is zero. You could have easily just looked at the fraction and made this safe bet.
    Last edited by colby2152; June 17th 2008 at 06:12 AM. Reason: 5^n + 5^n does not equal 10^n, how did I type that?
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  5. #5
    Super Member wingless's Avatar
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    Quote Originally Posted by colby2152 View Post
    Always try to compare the "top" and "bottom" of a fraction. If the "bottom" is always more than the "top" and either the "top" is decreasing and/or the "bottom" is increasing, then the limit of such a fraction is zero.

    "Top" = 2^n - 5^n
    "Bottom" = 3^n + 5^n

    Let's see if:

    "Top" < "Bottom"
    2^n - 5^n < 3^n + 5^n

    2^n -3^n<  + 10^n

    Note that 3^n > 2^n, so the left side yields a negative result. This inequality is always true.

    Therefore, we can safely say that the limit is zero. You could have easily just looked at the fraction and made this safe bet.
    2^n - 5^n < 3^n + 5^n

    2^n - 3^n < 2\cdot 5^n (also, this is not 10^n)

    You are not allowed to add 5^n to both sides. Doing such thing means,

    \lim_{x\to \infty} \frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f(x)+h(x)}{g(x)+h(x)} which is usually wrong.

    Also, 2^n - 5^n < 3^n + 5^n doesn't prove anything. LHS is approaching -\infty and RHS is approaching +\infty but who knows what their division will be?


    The easy way to do these questions is comparing the terms.

    1<2<...<n<x<2x<3x<...<x^2<x^2<x^n<...<x!<x^x

    The lesser terms will vanish.

    So,
    \lim_{n\to \infty} \frac{2^n - 5^n}{3^n+5^n}

    Those 5^n terms will grow much faster than 2^x and 3^x, making them meaningless. So we cross them out.

    \lim_{n\to \infty} \frac{\not 2^n - 5^n}{\not 3^n+5^n}

    \lim_{n\to \infty} \frac{- 5^n}{5^n}

    \boxed{-1}

    This was the term comparison method.


    The classic approach to these limits (as flyingsquirrel said) is dividing all terms by the biggest term.

    \lim_{n\to \infty} \frac{2^n - 5^n}{3^n+5^n}

    \lim_{n\to \infty} \frac{\frac{2^n}{5^n} - \frac{5^n}{5^n}}{\frac{3^n}{5^n}+\frac{5^n}{5^n}}

    We know that \lim_{x\to \infty}r^x = 0 for |r|<1. So \lim_{n\to \infty} \frac{2^n}{5^n} = \lim_{n\to \infty} \frac{3^n}{5^n} = 0

    \lim_{n\to \infty} \frac{0 - 1}{0+1} = \boxed{-1}
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by wingless View Post
    2^n - 5^n < 3^n + 5^n

    2^n - 3^n < 2\cdot 5^n (also, this is not 10^n)
    Thanks, I have no clue where that 10 came from... I fixed it!
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