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Math Help - 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  \alpha such that

    <br />
\lim_{x \to -2}\frac{3x^2+\alpha x + \alpha + 3}{x^2 + x - 2} <br />

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

    I tried first to factor the numerator into

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

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

    answer is
    \alpha = 15 limit = -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 \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

    \lim_ {x \to -2}<br />
 \frac<br />
{3\left(x+2\right)\left(x+3\right)}<br />
{\left(x+2\right)\left(x -1\right)}<br />
\Rightarrow<br />
\frac<br />
{3\left(x+3\right)}<br />
{\left(x -1\right)}<br />
\Rightarrow<br />
\frac<br />
 {3\left(-2+3\right)}<br />
 {\left(-2 -1\right)}<br />
\Rightarrow<br />
\frac<br />
 {3\left(1\right)}<br />
 {\left(-3\right)}<br />
\Rightarrow<br />
-1<br />
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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

     \lim_ {x \to -2}<br />
\frac<br />
{3\left(x+2\right)\left(x+3\right)}<br />
{\left(x+2\right)\left(x -1\right)}<br />
= \lim_ {x \to -2}<br />
\frac<br />
{3\left(x+3\right)}<br />
{\left(x -1\right)}<br />
= \lim_ {x \to -2}<br />
\frac<br />
{3\left(-2+3\right)}<br />
{\left(-2 -1\right)}<br />
= <br />
\frac<br />
{3\left(1\right)}<br />
{\left(-3\right)}<br />
=<br />
-1<br />
    Last edited by matheagle; January 2nd 2010 at 12:00 AM.
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