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Homework Help: Another math test question

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    i think I remember the question to be this:

    lim (y-3)/(y-3)(y+3)

    it was the first limit question and the easiest but im not sure if I did it right.

    2. Relevant equations

    3. The attempt at a solution

    so what I did was I cancelled the y-3's out to get:

    1/(y+3) and this is what I was unsure about is that the first step and after this I got:

    1/(3+3) = 1/6
  2. jcsd
  3. Oct 7, 2011 #2
    You're correct:


    \lim_{y \to \ 3} \frac{y-3}{(y-3)(y+3)} = \lim_{y \to \ 3 } \frac{1}{y+3} = \frac{1}{6}

  4. Oct 7, 2011 #3
    You have the correct answer, but it's probably a good idea to try to justify your last step. When is it true that [itex] \lim_{y \to a} f(y) = f(a) [/itex]? What can you say about the continuity of rational functions?
  5. Oct 7, 2011 #4


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    A crucial, and often overlooked property of limits is this: if f(x)= g(x) for all x except x= a, then [itex]\limit_{x\to a} f(x)= \limit_{x\to a}g(x)[/itex].

    For this problem, you can correctly say that
    [tex]\frac{y-3}{(y-3)(y+3)}= \frac{1}{y+3}[/tex]
    for all x except x= 3 and so the limits are the same.

    Actually, "what can you say about the continuity of rational functions?" is NOT a good question here because continuity also involves the value of the function as well as the limit. 1/(x+3) is continuous at x= 3, so its limit is 1/(3+3)= 1/6 while (x-3)/(x-3)(x+3) is NOT continuous at x= 3.
  6. Oct 7, 2011 #5
    This statement is true exactly because rational functions are continuous anywhere their denominator isn't zero, which is why I asked the question. I guess I should have mentioned the fact that there is a removable singularity at x=3, but the OP seemed to handle it appropriately.
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