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

1. Apr 22, 2008

### rsa58

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. Apr 22, 2008

### rsa58

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. Apr 22, 2008

### Dick

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. Apr 23, 2008

### rsa58

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. Apr 23, 2008

### rsa58

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. Apr 23, 2008

### Dick

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.