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Math Help - Help! Limits of functions

  1. #1
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    Help! Limits of functions

    Hi,

    Q1.

    lim (x = infinity) { (X^2 +5x - 6)/(4x^2 +x -8) }

    Q2.

    lim (x = infinity) { x^2/(x+5) - x }

    i know how to find limits of functions, but i dont know how to do it when x is infinity, please explain,

    thanks

    Moon
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  2. #2
    Member Mentia's Avatar
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    When you have a limit that looks like 0/0 or Infinity/Infinity thats when you use L'Hopital's rule.

    For the first one you can use the rule immediately. For the second one you have to get it in the form Infinity/Infinity so you get a common denominator and it turns into -5x/(x+5), and then use the rule.
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  3. #3
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    Hello, Moon!

    1)\;\;\lim_{x\to\infty}\frac{x^2 +5x - 6}{4x^2 +x -8}
    Divide top and bottom by x^2

    \lim_{x\to\infty}\frac{\dfrac{x^2}{x^2} + \dfrac{5x}{x^2} - \dfrac{6}{x}}{\dfrac{4x^2}{x^2} + \dfrac{x}{x^2} - \dfrac{8}{x^2}} \;=\; \lim_{x\to\infty}\frac{1 + \dfrac{5}{x} - \dfrac{6}{x^2}}{4 + \dfrac{1}{x} - \dfrac{8}{x^2}} \;=\;\frac{1+0-0}{4+0-0} \;=\;\frac{1}{4}




    2)\;\;\lim_{x\to\infty}\left(\frac{x^2}{x+5} - x\right)
    Combine the fractions: . \lim_{x\to\infty}\left(\frac{x^2-x(x+5)}{x+5}\right) \;=\;\lim_{x\to\infty}\frac{-5x}{x+5}


    Divide top and bottom by x

    . . \lim_{x\to\infty}\frac{\dfrac{-5x}{x}}{\dfrac{x}{x} + \dfrac{5}{x}} \;=\;\lim_{x\to\infty}\frac{-5}{1 + \dfrac{5}{x}} \;=\;\frac{-5}{1+0} \;=\;-5

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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    See this post, and read my generalization for limits of rational expressions (note this is just for \infty. You can make slight modifications for it as it goes to -\infty).
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