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Homework Help: Another Infinite Limit - Quite stuck

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{t \to \infty} \frac {5t+2}{t^{2}-6t+1} [/tex]

    2. Relevant equations



    3. The attempt at a solution

    Like the title says i am quite stuck, I don't see the first move to make, so if someone could point me in the right direction instead of telling me the answer it would be appreciated.

    I tried to factor the bottom trinomial but it wouldn't help out anyway, and fatoring out a t from the top and either a t or t^{2} from the bottom doesn't seem to get me any where. And the answer is suppose to be 0, according to the book.

    thanks again!
     
  2. jcsd
  3. May 18, 2010 #2
    You can start in a similar way to your last problem and factor out a t from both top and bottom. Look then at what the numerator and denominator tend to when t approaches infinity.
     
  4. May 18, 2010 #3
    ok well if you go at it that way it will be:

    [tex] \lim_{t \to \infty} \frac {t(5+2t^{-1})}{t(t-6+t^{-1})} [/tex]

    [tex] \lim_{t \to \infty} \frac {5+2t^{-1}}{t-6+t^{-1}} [/tex]

    which is saying that when:

    [tex] H= \infty \;\;\;\;\;and\;\;\;\;\; \epsilon=\frac{1}{H} = infinitesimal [/tex]

    it is:

    [tex] \frac {5 + \epsilon}{H-6+\epsilon} [/tex]

    which is basically 5 divided by infinity, I don't see why that that equals 0

    according to my notes an a constant number divided by an infinite number is going to be an infinitesimal number, is that why the limit is 0? because

    [tex] \frac {5}{H} = \epsilon [/tex]

    and epsilon is going to be infinitely close to 0?
     
  5. May 18, 2010 #4

    Mark44

    Staff: Mentor

    The larger H gets, the smaller epsilon gets, so yes, 5/H can be made arbitrarily close to zero.
     
  6. May 18, 2010 #5
    ok thanks for the help again guys
     
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