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Math Help - Limits

  1. #1
    Junior Member
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    Limits

    I'm having some troubles reducing this.

    I need to find the limit of x^3 + 2x^2 - 5x - 6 / (x+ 3) where x approches -3?

    ive forgotten how to factor x^3 + 2x^2 - 5x - 6

    help would be much appreciated
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  2. #2
    Member eXist's Avatar
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    Because the limit approaches \frac{0}{0} when you substitute -4 and \infty when you substitute \infty you can apply Hospital's rule:

    \lim_{x\to a} f(x)=\frac{g(x)}{h(x)}

    if \lim_{x\to a} g(x) = \lim_{x\to a} h(x) = 0

    then \lim_{x\to a} f(x)=\frac{g(x)}{h(x)} = \lim_{x\to a} f(x)=\frac{g'(x)}{h'(x)}

    So:
    \lim_{x\rightarrow-3}\frac{x^3 + 2x^2 - 5x - 6}{x+ 3} =<br />
\lim_{x\rightarrow-3}\frac{\frac{d}{dx}(x^3 + 2x^2 - 5x - 6)}{\frac{d}{dx}(x+ 3)} =<br />
\lim_{x\rightarrow-3}\frac{3x^2 + 4x - 5}{1} = 10<br />
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  3. #3
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    x^3+2x^2-5x-6 = (x+3)(x-2)(x+1)

    lim_{x\to-3}\frac{x^3+2x^2-5x-6}{x+3} = \lim_{x\to-3} (x-2)(x+1)

    = \lim_{x\to-3} x^2 -x -2 = 10
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  4. #4
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    Quote Originally Posted by foreverbrokenpromises View Post
    I'm having some troubles reducing this.

    I need to find the limit of x^3 + 2x^2 - 5x - 6 / (x+ 3) where x approches -3?

    ive forgotten how to factor x^3 + 2x^2 - 5x - 6

    help would be much appreciated
    The only reason you want to factor that is because both it and the denominator become 0 when x= -3. That means that x-(-3)= x+ 3 is a factor. Now just do a "long division" to find the other factor: x goes into x^3 x^2 times. Multiplying x^2 by x+ 3 we get \begin{array}{cccc}x^3& +2x^2 & -5x& -6 \\ x^2 & +3x^2 & 0 & 0\\0 & -x^2& -5x& -6\end{array}

    Now, x goes into -x^2 -x times. Multiplying -x by x+ 3 we get \begin{array}{ccc}-x^2 & -5x& -6 \\-x^2 & -3x& -6 \\ 0 & -2x & -6\end{array}.

    Finally, x goes into -2x -2 times and, multiplying -2 by x+ 3 we get \begin{array}{cc}-2x & -6 \\ -2x & -6 \\ 0 & 0\end{array}

    x^3+ 2x^2- 5x- 6= (x+ 3)(x^2- x- 2).
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