1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How would I solve this limit?

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data

    (2+y^2)/sqrt(4+y^4) as y goes to positive infinity

    2. Relevant equations

    3. The attempt at a solution

    I divided it out by y^2 to get

    lim y-> inf+ (2+y^2)/y^2 = 1 and the bottom lim y -> inf+ sqrt(4+y^4)/y^2 I can't solve..

    by the way, I can't use L'hopitals rule

  2. jcsd
  3. Dec 8, 2009 #2
    Split them into 2.

    1. [tex] 2 + y^2[/tex] --> inf as y --> inf
    2. [tex] 1/sqrt(4 + y^4) [/tex] --> 0 as y -- inf

    So you are having an inf / 0 situation.
  4. Dec 8, 2009 #3

    I know, that's why divided it out by y^2.. so the question really comes down to solving

    lim y -> inf+ sqrt(4+y^4)/y^2
    thanks for reading
  5. Dec 8, 2009 #4
    First of all, notice that (2+y^2)^2=(4+4y^2+y^4)>(4+y^4) for y positive.
    This means (2+y^2)>sqrt(4+y^4)... which means that the limit you're looking for is either greater than or equal to 1 or infinite (why?).

    As for evaluating the actual limit, you might find it easier to first deal with the ratio (4+4y^2+y^4)/(4+y^4)
  6. Dec 8, 2009 #5


    Staff: Mentor

    Factor y^2 out of both the numerator and denominator and then take the limit.
    [tex]\frac{2 + y^2}{\sqrt{4 + y^4}}~=~\frac{y^2(2/y^2 + 1)}{y^2\sqrt{4/y^4 + 1}}[/tex]

    Can you finish it?
  7. Dec 8, 2009 #6
    thanks, I can see that for large y, it is 1
  8. Dec 8, 2009 #7


    Staff: Mentor

    Yes. And you can see that in the original expression, as well. For large y, the numerator is pretty close to y^2, and the denominator is also pretty close to y^2, since the square root of y^4 is y^2. The constants in the top and bottom are less and less significant the larger y gets.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook