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Math Help - Finding limits at infinity - help on a couple problems

  1. #1
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    Finding limits at infinity - help on a couple problems

    I managed to get the first couple questions in the section right..but as most of you know they get harder the farther you go. Currently these are the two I am stuck on. If somebody could show me how to solve them I can hopefully get the hang of the rest of them.

    Decide whether the limits in the following problems exist. If a limit exist, find its value. The set of numbers should be next to the limit x approaching infinity but it came out sloppy so I just put them below.

    Problem 1
    lim
    x-->infinity

    x^2 + 6x + 8
    x^3 + 2x + 1




    Problem 2
    lim
    x--> - infinity

    ( 9/x^4 + 1/x^2 - 3)


    Their are parentheses surrounding the entire problem except the limit(In other words from the 9/x^4 to the -3) I didn't put them as a fraction with an underline like problem 1 because they are seperate and it came out too sloppy, but you get the idea, it's 9 over x^4 plus 1 over x^2 minus 3

    Thanks in advance for the help!
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  2. #2
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    Quote Originally Posted by Mathamateur View Post

    Problem 1
    lim
    x-->infinity

    x^2 + 6x + 8
    x^3 + 2x + 1
    By dividing throw by the largest exponent x^3 these functions agree for sufficiently large x.

    Thus,
    \lim_{x\to \infty} \frac{\frac{1}{x}+\frac{6}{x^2}+\frac{8}{x^3}}{1+\  frac{2}{x^2}+\frac{1}{x^3}}=\frac{0+0+0}{1+0+0}=0


    Problem 2
    lim
    x--> - infinity

    ( 9/x^4 + 1/x^2 - 3)
    !
    That is,
    0+0-3=-3
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  3. #3
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    Quote Originally Posted by Mathamateur View Post
    Problem 1
    lim
    x-->infinity

    x^2 + 6x + 8
    x^3 + 2x + 1
    \lim_{x \to \infty} \frac{x^2+6x+8}{x^3+2x+1}

    Divide the numerator and denominator by \frac{1}{x^2}:

    = \lim_{x \to \infty}\frac{1 + \frac{6}{x} + \frac{8}{x^2}}{x + \frac{2}{x} + \frac{1}{x^2}}

    Taking the limit of the numerator denominator we see that \lim_{x \to \infty}\frac{1}{x} and \lim_{x \to \infty}\frac{1}{x^2} are both 0, so

    \lim_{x \to \infty} \frac{x^2+6x+8}{x^3+2x+1} \to \lim_{x \to \infty}\frac{1}{x} \to 0

    -Dan
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    So does the limit not exist, or is the limit zero?
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    Forum Admin topsquark's Avatar
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    Why would you be thinking it wouldn't exist?

    The limit in the first is 0.
    The limit in the second is -3.

    -Dan
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    Quote Originally Posted by Mathamateur View Post
    So does the limit not exist, or is the limit zero?
    Whenever a limit is equal to a number it exists.
    Otherwise, it does not exist.
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