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Math Help - evaluating limits. (part deux)

  1. #1
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    evaluating limits. (part deux)

    Evaluate the limit, if it exists.



    im not sure how to get this problem going, I try factoring the numerator and get x(x-9) but theres nothing in the denominator (that i can spot, anyways) that also gets me (x-9). Would I have to use a conjugate?

    Thank you.
    Evan
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  2. #2
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    Quote Originally Posted by Evan.Kimia View Post
    Evaluate the limit, if it exists.



    im not sure how to get this problem going, I try factoring the numerator and get x(x-9) but theres nothing in the denominator (that i can spot, anyways) that also gets me (x-9). Would I have to use a conjugate?

    Thank you.
    Evan
    Do you know how to factorise Quadratic Trinomials? This is the second question you've asked which is easily solved through factorisation...

    x^2 - 8x - 9.


    You need to think of two numbers that multiply to be -9 and add to be -8.

    They are -9 and 1.

    So

    x^2 - 8x - 9 = x^2 - 9x + 1x - 9

     = x(x - 9) + 1(x - 9)

     = (x - 9)(x + 1).


    Therefore

    \frac{x^2 - 9x}{x^2 - 8x - 9} = \frac{x(x - 9)}{(x - 9)(x + 1)}.


    Go from here.
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  3. #3
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    I didnt know how to factor quadratic trinomials but with your input and a google search I do now. Before this ive just been guessing the 2 numbers :/. Thanks for making my life easier.
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