# Limit in parametric form

1. Jan 30, 2010

### talolard

1. The problem statement, all variables and given/known data
Calculate $$lim{t-> + \infty} ( \frac{a^t + b^t)}{2}) ^ {1/t}$$

2. Relevant equations

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Jan 30, 2010
2. Jan 30, 2010

### 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

3. Jan 30, 2010

### talolard

I dont see how that helps/ I get $$\frac {ln(a^t+b^t)/2}{t}$$ which seems to have an infinite limit

4. Jan 30, 2010

### 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

5. Jan 30, 2010

### 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 recieve infinity / infinity everttime.
Thanks

6. Jan 30, 2010

### 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