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Math Help - limits, finding value of b

  1. #1
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    limits, finding value of b

    Is there a number b such that \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.
    -1^2-(-1)-2=0
    So for the numerator I am going to plug in -1 as well and solve for b
    2(-1)^2+b(-1)+3b=0
    That gives me b=-1
    And \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.

    -1^2-(-1)-2=0

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

    2(-1)^2+b(-1)+3b=0

    That gives me

    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
    If you want to find the limit as 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 (x+1) factor,

    as (x+1)=0 if x=-1

    \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 (x+1) as one of those factors,
    we will be able to find the limit...

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

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

    Hence, comparing terms we have

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

    Hence

    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\lim_{x \to -1}\frac{(x+1)(2x+c)}{(x+1)(x-2)}=\frac{2x+c}{x-2} evaluated at x=-1
    Last edited by Archie Meade; September 26th 2010 at 10:52 AM. Reason: added final lines for completion
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