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Math Help - Really Simple Limit Question

  1. #1
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    Really Simple Limit Question

    I was just wandering if we had say  \frac{(x^2 - x -6)^2}{(x+2)^2}, if we had to find the limit of that as it approached -2 for example.
    Could I cancel both top and bottom by (x + 2)^2, or would that not be correct since (x+2) = 0 when x = -2
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  2. #2
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    ummm im not surei know u could write it as \frac{(x - 3)^2(x + 2)^2}{(x + 2)^2} but im not sure if ur allowed to cancel since the bottom tends to 0.
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  3. #3
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    Hello,
    Quote Originally Posted by Lipticboven View Post
    ummm im not surei know u could write it as \frac{(x - 3)^2(x + 2)^2}{(x + 2)^2} but im not sure if ur allowed to cancel since the bottom tends to 0.
    You can simplify, and that's the principle of these limits.

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

    So, firstly without working with limits, \frac{(x^2-x-6)^2}{(x+2)^2}=\frac{(x-3)^2(x+2)^2}{(x+2)^2}=(x-3)^2
    You can simplify, because you don't know yet top and bottom are 0. There's no problem.

    Now, take the limit as x tends to 2.
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