Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Limit problem (high school calculus) help?

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  • 1 month ago

    You'd have 3x^2 / √(x^2) if you deal with only the highest exponents of x.  For all non-negative x, √(x^2) = x, so since x approaches infinity here you'd have 3x^2 / x, which simplifies to 3x.  This increases forever without limit (or, rather, the limit is infinity), so the limit of the whole expression is infinity.

    Because you have a limit of infinity, there would be no horizontal asymptote; for that to be the case the limit would have to approach a certain value.

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  • Vaman
    Lv 7
    1 month ago

    When x is very large, then you can neglect x. The equation becomes

    3x^2/x=3x. The limit becomes infinity.

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  • (3x^2 + 2x) / sqrt(x^2 - 2x) =>

    x * (3x + 2) / (sqrt(x^2 * (1 - 2/x))) =>

    x * (3x + 2) / (x * sqrt(1 - 2/x)) =>

    (3x + 2) / sqrt(1 - 2/x)

    x goes to infinity

    (3 * inf + 2) / sqrt(1 - 2/inf) =>

    inf / sqrt(1 - 0) =>

    inf / sqrt(1) =>

    inf / 1 =>

    inf

    As x goes to infinity, the limit goes to infinity.  What does it say about the existence of horizontal asymptotes?  Beats me, since there aren't any as x approaches infinity.

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