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The limit of (xn + yn)^(1/n)

  1. Apr 22, 2008 #1
    1. The problem statement, all variables and given/known data

    3. (10 points) Show that if {xn}, {yn} are two sequences of positive real
    numbers then
    lim (xn +yn)^(1/n) is the max of lim(xn)^(1/n) and lim(yn)^(1/n)

    provided these limits exist.

    2. Relevant equations
    Does anyone have any nice inqualities or tricks?


    3. The attempt at a solution
     
  2. jcsd
  3. Apr 22, 2008 #2
    i got the following:
    set sn = (xn + yn)^(1/n). then xn + yn <= (epsilon + |x|)^n + (epsilon + |y|)^n

    <= (2epsilon + |x| + |y|)^n ..... sn - |x| - |y| <= 2epsilon.

    does this mean that the limit is the sum of the two limits for xn and yn?
     
  4. Apr 22, 2008 #3

    Dick

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    If (xn)^1/n -> X and (yn)^(1/n) -> Y then roughly speaking xn is like X^n and yn is like Y^n. So practice by figuring out why (X^n+Y^n)^(1/n) approaches max(X,Y). If you need a formal proof then you'll have to figure out how to quantify 'is like' in terms of epsilon.
     
  5. Apr 23, 2008 #4
    uh, that's what i did though. i said if |(xn)^(1/n) - x | < epsilon

    then by triangle inequality we have |xn^(1/n)| < epsilon + |x|

    but doesn't this give |xn| < (epsilon + |x|)^n the same hold for yn.

    xn + yn < (epislon + |x|)^n + (epsilon +|y|)^n < (epsilon + |x| + epsilon + |y|)^n

    and |(xn + yn)^(1/n) - (x + y)| < 2epsilon.
     
  6. Apr 23, 2008 #5
    i don't see how (x^n + y^n)^(1/n) ----> max (x, y).

    i mean, look, x^n + y^n <= (x+y)^n since x and y are non negative.

    then sn <= x+y. so it is bounded. but isn't this bound also the limit by the above epsilon argument?
     
  7. Apr 23, 2008 #6

    Dick

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    It's a good idea to get this out of the way first. In (x^n+y^n)^(1/n), suppose x>y. Factor the x's outside of the root. What happens? If y>x, factor the y's out.
     
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