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Thread: Limit question

  1. #1
    Member Vinod's Avatar
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    Question Limit question

    Determine the value of a so that $f(x) = \frac{x^2+ ax + 5}{x+1} $has a slant asymptote y = x + 3
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  2. #2
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    Re: Limit question

    You want the following limit to exist and equal zero:

    $$\lim_{x \to \infty} [f(x) - (x+3)] = 0$$

    Solve for $a$. You want the numerator to be a constant.
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    Re: Limit question

    $$f(x) = \frac{x^2+ ax + 5}{x+1} = (x+3) + \frac{b}{x+1} \, \text{($b$ constant)}$$
    Last edited by Archie; Apr 11th 2018 at 11:19 AM.
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  4. #4
    Member Vinod's Avatar
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    Re: Limit question

    So a=4 and b=2
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    Re: Limit question

    Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$.
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    Re: Limit question

    Quote Originally Posted by HallsofIvy View Post
    Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$.
    Small typo:

    For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$.
    Last edited by SlipEternal; Apr 12th 2018 at 05:21 AM.
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    Re: Limit question

    Do you mean "for X not equal to -1"?
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    Re: Limit question

    Quote Originally Posted by Archie View Post
    Do you mean "for X not equal to -1"?
    Lol, good point. Fixed.
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    Re: Limit question

    Now it says "X not equal to 1", not "X not equal to -1"
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  10. #10
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    Re: Limit question

    Quote Originally Posted by SlipEternal View Post
    Small typo:

    For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$.
    Yes. Thank you and Archie.
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