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Homework Help: More limits, just a simple question this time

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    I just have a question about these two questions:

    [tex] \lim_{y \to 3^{+}} \frac {y+1}{(y-2)(y-3)} [/tex]

    and

    [tex] \lim_{y \to 3} \frac {y+1}{(y-2)(y-3)} [/tex]

    the solution to the first problem is infinity and the solution to the second is does not exist, is this because the first one is is approach 3 from the positive side so its basically 3.000000000000...1 and therefore infinity and the second is at 3 so the denominator is 0? If i am wrong please clarify this issue for me.

    thanks!
     
  2. jcsd
  3. May 19, 2010 #2
    The first one is infinity because you're essentially dividing by a very small positive number since you're approaching 3 from the right. When applying the limit, you will get something like 4/(1 x 0^+), meaning your dividing 4 by a very very small number therefore the limit is infinity.

    The second one does not exist because if you take the one sided limits, you will get +infinity when you approach 3 from the right and -infinity when you approach 3 from the left. Since the limits from both sides don't match, the limit as x approaches 3 DNE.
     
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