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Math Help - Stuck on easy limit

  1. #1
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    Stuck on easy limit

    Hi all, stuck on the following limit:
    \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
    I have managed to factor it out to:
    \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
    Thanks
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  2. #2
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    Quote Originally Posted by Oiler View Post
    Hi all, stuck on the following limit:
    \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
    I have managed to factor it out to:
    \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
    Thanks
    \dfrac{t^3 + 3t^2 - 12t + 4}{t^3 - 4t} = \dfrac{(t-2)(t^2+5t-2)}{t(t-2)(t+2)}
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  3. #3
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    Quote Originally Posted by Oiler View Post
    Hi all, stuck on the following limit:
    \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}
    I have managed to factor it out to:
    \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2
    Thanks
    You could try

    \displaystyle\frac{t^3+3t^2-12t+4}{t^3-4t}=\frac{\left(t^3-4t\right)+3t^2-8t+4}{t^3-4t}=1+\frac{3t^2-8t+4}{t^3-4t}

    =\displaystyle\ 1+\frac{(3t-2)(t-2)}{t\left(t^2-4\right)}=1+\frac{(3t-2)(t-2)}{t(t+2)(t-2)}
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