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Math Help - Limit problem

  1. #1
    Junior Member
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    Limit problem

    lim(as x approaches -9) of (x^2-17x+72)/(x+9)
    my first step was to factor the top to. {(x-8)(x-9)}/(x+9)

    Usually at this point you should be able to cancel out a factor but I think I am going at this problem wrong. Any ideas?
    Thanks
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  2. #2
    MHF Contributor

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    No, there is no factor in the numerator that is the same as the factor in the denominator so there is way to cancel. Notice thaty if you simply put x= -9 in the fraction you get \frac{(-9)^2- 17(-9)+ 72}{-9+ 9}= \frac{81+ 153+ 72}{0}= \frac{305}{0} which does not exist ("is infinity"). The limit does not exist.

    It is only when you get "0/0" that you can factor and possibly get a limit.

    If you expect this to have a limit, you might want to check that you copied the problem correctly!
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  3. #3
    Junior Member
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    Hmm that is weird. I originally came up with DNE but when I plugged it into the online HW assignment it marked it as incorrect but now i just went back in and it went through. I dunno what happened. Thanks for the explanation!
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