What is the Limit of (a^t + b^t)/2^(1/t) as t approaches infinity?

[talolard]

Homework Statement

Calculate lim{t-> + \infty} ( \frac{a^t + b^t)}{2}) ^ {1/t}

Homework Equations

The Attempt at a Solution

lim _{t-> + \infty} ( \frac{a^t + b^t)}{2}) ^ {1/t} = lim_{t-> + \infty} ( a^t ( \frac{1 + b^t/a^t)}{2}) ^ {1/t} = lim_{t-> + \infty} ( a ( \frac{1 + (b/a)^t)}{2}) ^ {1/t} = lim_{t-> + \infty} ( a ( \frac{1 + (b/a)^t)^{1/t} }{2}) ^ {1/t}) = lim_{t-> + \infty} ( a ( {1 + (b/a)^t)^{1/t}=a

Because 1\t aproacges 0 as t aproaches infinity.
Is this ok?
thanks
Tal

 

[torquil]

Maybe you could try to calculate the limit of the logarithm of this expression instead? Find the limit, and then transform it back using the inverse log.

Torquil

 

[talolard]

I don't see how that helps/ I get \frac {ln(a^t+b^t)/2}{t} which seems to have an infinite limit

 

[torquil]

The denominator has an infinite limit, and the numerator will have either an infinite positive or negative limit depending a bit on the values of a and b. Or I guess undefined if a and/or b are allowed to be complex numbers.

There are some standard techniques that are used to deal with this case when both the numerator and denominator diverge towards infinity.

Torquil

 

[talolard]

I'm assuming you are talking about Le'Hospital (Pardon the spelling, english is not my native language, we call it Lupital).
So taking it that way i have
lim_{t-> \infty} ln( \frac {a^t + b^t}{2})^(1/t)= lim_{t-> \infty} \frac {(ln(a^t + b^t)/2)}{t} = \frac {ln(a)a^t+ln(b)b^t}{2(a^t+b^t)}
at which point i get stuck. Because i will receive infinity / infinity everttime.
Thanks

 

[torquil]

Yes, but now your can e.g. divide by a^t both over and under, to get expressions like (b/a)^t. And you can define x := (b/a)^t. E.g., the limit t -> infty would be the same as x -> infty if b/a>1. After that you can use L'Hopital again on the x-variable to get the answer.

Of course, maybe it would have been possible to use the x := (b/a)^t variable all the way from the beginning, I didn't think of that at first.

Torquil