What are some tricks for finding the limit of (xn + yn)^(1/n)?

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Homework Help Overview

The problem involves finding the limit of the expression (xn + yn)^(1/n) where {xn} and {yn} are sequences of positive real numbers. The original poster seeks inequalities or tricks to demonstrate that this limit is the maximum of the limits of (xn)^(1/n) and (yn)^(1/n) if those limits exist.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various approaches to understanding the limit, including inequalities and the triangle inequality. Some explore the relationship between the sequences and their limits, while others question the implications of bounding the expression.

Discussion Status

The discussion includes attempts to clarify the reasoning behind the limit and the relationships between the sequences. Some participants have offered insights into bounding the expression, while others are exploring the implications of their findings without reaching a consensus.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share. There is an ongoing examination of assumptions related to the behavior of the sequences as n approaches infinity.

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Homework Statement



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.

Homework Equations


Does anyone have any nice inqualities or tricks?


The Attempt at a Solution

 
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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?
 
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.
 
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.
 
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?
 
rsa58 said:
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?

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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