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

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In summary: This gives you (x^n+y^n)^(1/n) which is greater than (x^n+y^n) because the y's have been factored out.
  • #1
rsa58
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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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  • #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?
 
  • #3
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.
 
  • #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.
 
  • #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?
 
  • #6
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.
 

What is the limit of (xn + yn)^(1/n)?

The limit of (xn + yn)^(1/n) as n approaches infinity is equal to the maximum of the two terms, xn and yn.

How can the limit of (xn + yn)^(1/n) be calculated?

The limit of (xn + yn)^(1/n) can be calculated using the L'Hopital's rule or by factoring out the largest exponent of n from the expression.

What happens to the limit of (xn + yn)^(1/n) when x and y have opposite signs?

If x and y have opposite signs, the limit of (xn + yn)^(1/n) will approach zero as n approaches infinity, since the smaller term will dominate the expression.

What if x and y have the same sign?

If x and y have the same sign, the limit of (xn + yn)^(1/n) will approach the value of the larger term as n approaches infinity.

What is the significance of the limit of (xn + yn)^(1/n)?

The limit of (xn + yn)^(1/n) is significant in understanding the behavior of sequences and series, and can be used to determine the convergence or divergence of certain mathematical expressions.

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