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Homework Help: 1st year calc, limits

  1. May 12, 2006 #1
    1 More Question
    It says Calculate the following limits:
    lim x-> -3 ((1/x+3)+6/(x^2-9)) So what i did was changed x^2-9 into (x-3)(x+3) and then made that the common denominator so I got
    1(x-3) + 6/(x-3)(x+3) then the (x-3)'s cancel and your left with 6/(x+3) but then I plug int he -3 and I get 6/0 which is undefined, is that the answer that it doesnt exist or am I doing something wrong? THANKS
  2. jcsd
  3. May 12, 2006 #2
    you're looking for [tex]lim_{\substack{x\rightarrow -3} \frac{1}{x+3} + \frac{6}{x^2-9} [/tex]...you are correct to look for a common denominator...but you made a mistake at the point where you cancel out terms. What is 1(x-3) + 6 ? :wink:
    Last edited: May 12, 2006
  4. May 12, 2006 #3

    Ohh ok yea I get it so 1(x-3)+6 is x-3+6 so x+3 then the x+3's cancel out and you get 1/(x-3) which is 1/-6... THANKS!!
  5. May 12, 2006 #4
    cool! :smile:
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