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Math Help - Slant Asymptote

  1. #1
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    Slant Asymptote

    Given the function f(x)=\frac{3x^4+4}{x^3-3x} I have found, by long division, that there is a slant asymptote, y=3x.
    Further, I wanted to find the same result by the classic limit definition.

    \lim_{x \mapsto \infty}\left [ \frac{3x^4+4}{x^3-3x} \right ]

    \lim_{x \mapsto \infty}\left [ \frac{x^4(1+\frac{4}{x^4})}{x^4(\frac{1}{x}-\frac{3}{x^3})} \right ]

    \lim_{x \mapsto \infty}\left [ \frac{1+\frac{4}{x^4}}{\frac{1}{x}-\frac{3}{x^3}} \right ]=\frac{1+0}{0-0}

    I ended up in an indeterminate form.The result should had be the same!What's wrong?
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  2. #2
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    Re: Slant Asymptote

    \frac{1}{0} is not an indeterminate form.
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  3. #3
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    Re: Slant Asymptote

    I made a mistake in the simplification.The final result is \frac{3}{0}. The problem is division by zero!
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    Re: Slant Asymptote

    Quote Originally Posted by joaofonseca View Post
    I made a mistake in the simplification.The final result is \frac{3}{0}. The problem is division by zero!
    which means the limit does not exist.
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  5. #5
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    Re: Slant Asymptote

    The limit exist!Look at the graph:

    Slant Asymptote-slant.jpg
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  6. #6
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    Re: Slant Asymptote

    I'm afraid you are confused ... the limit as x \to \infty will determine a horizontal asymptote, but not a slant asymptote.

    After you do the long division, you get an expression of the form y = mx + b + r(x), where r(x) is a remainder that tends to 0 as x \to \infty , telling you that the function value approaches the line y = mx+b
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