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Homework Help: 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.
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