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Math Help - horizontal asymptot

  1. #1
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    horizontal asymptot

    how to finf the horizontal asymptot of
    f(x) = x^2-1/x^2+x-2
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  2. #2
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    [tex]


    It's just a matter of observation. If the power of the numerator is the same as the power of the denominator, then the asymptote is the line y=\frac{a_{n}}{b_{n}}. Which are the coefficients of the leading terms.
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  3. #3
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    Hello, wutever

    Find the horizontal asymptote of: . f(x) \:= \:\frac{x^2-1}{x^2+x-2}
    The traditional approach . . .

    Find: . \lim_{x\to\infty} f(x)

    If the limit has a finite value a, the horizontal asymptote is: . y = a


    We have: . \lim_{x\to\infty}\,\frac{x^2-1}{x^2+x-2}

    Divide top and bottom by the highest power of x in the denominator:

    . . \lim_{x\to\infty}\,\frac{\frac{x^2}{x^2} - \frac{1}{x^2}}{\frac{x^2}{x^2} + \frac{x}{x^2} - \frac{2}{x^2}} \;=\;\lim_{x\to\infty}\,\frac{1 - \frac{1}{x^2}}{1 + \frac{1}{x} - \frac{1}{x^2}} \;=\;\frac{1 - 0}{1 + 0 - 0} \;=\;1


    Therefore, the horizontal asymptote is: . y \,=\,1

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  4. #4
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    thx for the reply
    i have another one (sqr root of 9x^2+5x-2) / 2x^3+1
    how to find its horizontal asymptot..
    i have the answer which is (sqr root of 9) / 2
    so is it a rule that the asymptot is the cooeficent of x in the denominator over the cooficent of x in the numenerator ???
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wutever View Post
    thx for the reply
    i have another one (sqr root of 9x^2+5x-2) / 2x^3+1
    clarify, is this f(x) = \frac {\sqrt{9x^2 + 5x - 2}}{2x^3 + 1} ?

    if that's the case, your answer is wrong

    how to find its horizontal asymptot..
    exactly the same way, find \lim_{x \to \infty}f(x) and \lim_{x \to - \infty} f(x)

    i have the answer which is (sqr root of 9) / 2
    so is it a rule that the asymptot is the cooeficent of x in the denominator over the cooficent of x in the numenerator ???
    no, only when the highest power in the numerator is the SAME as the highest power in the denominator, then the answer is the ratio of the two coefficients, otherwise, not necessarily
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  6. #6
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    am sorry i wrote the power wrong, the right powers is 6 instead of 2 in the denominator
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wutever View Post
    am sorry i wrote the power wrong, the right powers is 6 instead of 2 in the denominator
    in that case, the answer would be that you have 2 horizontal asymptotes: \pm \frac 32

    we want \lim_{x \to \infty}f(x) and \lim_{x \to - \infty}f(x) where f(x) = \frac {\sqrt{9x^6 + 5x - 2}}{2x^3 + 1}

    now as x goes to infinity, only the highest powers matter, since doing trivial things like subtracting 2 or adding 1 would be inconsequential "at" infinity. and so would adding 5x when it is contending with 9x^6.

    thus: \lim_{x \to \infty} \frac {\sqrt{9x^6 + 5x - 2}}{2x^3 + 1} = \lim_{x \to \infty} \frac {\sqrt{9x^6}}{2x^3}

    = \lim_{x \to \infty} \frac {\sqrt{9x^4x^2}}{2x^3}

    = \lim_{x \to \infty} \frac {3x^2 \sqrt{x^2}}{2x^3}

    = \lim_{x \to \infty} \frac {3|x|}{2x} ........since |x| = \sqrt{x^2}

    = \lim_{x \to \infty} \frac 32 \frac {|x|}{x}

    so we have: \frac 32 \frac {|x|}x \to \left \{ \begin{array}{lr}  ~~\frac 32 & \mbox{ if } x \ge 0 \mbox{ that is, } x \to \infty \\ & \\  - \frac 32 & \mbox{ if } x < 0 \mbox{ that is, } x \to - \infty \end{array} \right.
    Last edited by Jhevon; October 23rd 2007 at 01:28 PM. Reason: fixed LaTex code
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