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Thread: find a such that and the limit

  1. #1
    Super Member bigwave's Avatar
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    Cool find a such that and the limit

    Is there a number $\displaystyle \alpha$ such that

    $\displaystyle
    \lim_{x \to -2}\frac{3x^2+\alpha x + \alpha + 3}{x^2 + x - 2}
    $

    exists? If so find the value of
    $\displaystyle \alpha$ and the value of the Limit

    I tried first to factor the numerator into

    $\displaystyle \left(x-1\right)\left(x+2\right)$

    thinking this would reveal how to factor the numerator
    to find
    $\displaystyle \alpha$ but nothing became obvious.

    answer is
    $\displaystyle \alpha = 15$ limit $\displaystyle = -1$
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  2. #2
    MHF Contributor matheagle's Avatar
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    you need the numerator to be 0 when x=-2
    Inserting -2 for x does make $\displaystyle \alpha=15$
    THEN factor the numerator and cancel the x+2 terms.
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  3. #3
    Super Member bigwave's Avatar
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    so then

    $\displaystyle \lim_ {x \to -2}
    \frac
    {3\left(x+2\right)\left(x+3\right)}
    {\left(x+2\right)\left(x -1\right)}
    \Rightarrow
    \frac
    {3\left(x+3\right)}
    {\left(x -1\right)}
    \Rightarrow
    \frac
    {3\left(-2+3\right)}
    {\left(-2 -1\right)}
    \Rightarrow
    \frac
    {3\left(1\right)}
    {\left(-3\right)}
    \Rightarrow
    -1
    $
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  4. #4
    MHF Contributor matheagle's Avatar
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    I like to see the limit signs

    Quote Originally Posted by bigwave View Post
    so then

    $\displaystyle \lim_ {x \to -2}
    \frac
    {3\left(x+2\right)\left(x+3\right)}
    {\left(x+2\right)\left(x -1\right)}
    = \lim_ {x \to -2}
    \frac
    {3\left(x+3\right)}
    {\left(x -1\right)}
    = \lim_ {x \to -2}
    \frac
    {3\left(-2+3\right)}
    {\left(-2 -1\right)}
    =
    \frac
    {3\left(1\right)}
    {\left(-3\right)}
    =
    -1
    $
    Last edited by matheagle; Jan 2nd 2010 at 12:00 AM.
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