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Thread: limits, finding value of b

  1. #1
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    limits, finding value of b

    Is there a number b such that $\displaystyle \lim_{x \to -1} \frac{2x^2+bx+3b}{x^2-x-2}$ exists? If so, find the value of b and the value of the limit.
    I'm just going to plug -1 into the denominator and that gives 0.
    $\displaystyle -1^2-(-1)-2=0$
    So for the numerator I am going to plug in -1 as well and solve for b
    $\displaystyle 2(-1)^2+b(-1)+3b=0$
    That gives me $\displaystyle b=-1$
    And $\displaystyle \lim_{x \to -1} \frac{2x^2-x-3}{x^2-x-2}=\frac{5}{3}$

    Am I doing it right? Thanks
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  2. #2
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    Quote Originally Posted by iamanoobatmath View Post
    I'm just going to plug -1 into the denominator and that gives 0.

    $\displaystyle -1^2-(-1)-2=0$

    So for the numerator I am going to plug in -1 as well and solve for b

    $\displaystyle 2(-1)^2+b(-1)+3b=0$

    That gives me

    $\displaystyle b=-1$

    And

    $\displaystyle \displaystyle\lim_{x \to -1} \frac{2x^2-x-3}{x^2-x-2}=\frac{5}{3}$

    Am I doing it right? Thanks
    If you want to find the limit as $\displaystyle x \to -1$

    you must resolve your "divide-by-zero" problem.
    Therefore, try to factor the denominator to see if we can deal with an $\displaystyle (x+1)$ factor,

    as $\displaystyle (x+1)=0$ if $\displaystyle x=-1$

    $\displaystyle \displaystyle\frac{2x^2+bx+3b}{x^2-x-2}=\frac{2x^2+bx+3b}{(x+1)(x-2)}$

    Now, if we can factor the numerator, with $\displaystyle (x+1)$ as one of those factors,
    we will be able to find the limit...

    $\displaystyle 2x^2+bx+3b=(x+1)(2x+c)$

    $\displaystyle \Rightarrow\ 2x^2+bx+3b=2x^2+cx+2x+c=2x^2+(c+2)x+c$

    Hence, comparing terms we have

    $\displaystyle c+2=b,\;\;\;c=3b$

    Hence

    $\displaystyle 3b+2=b\Rightarrow\ 2b=-2\Rightarrow\ b=-1$

    You can then evaluate the limit if you like.

    The limit is evaluated by simplifying the fraction

    $\displaystyle \displaystyle\lim_{x \to -1}\frac{(x+1)(2x+c)}{(x+1)(x-2)}=\frac{2x+c}{x-2}$ evaluated at $\displaystyle x=-1$
    Last edited by Archie Meade; Sep 26th 2010 at 09:52 AM. Reason: added final lines for completion
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