Limit as x approaches infinity of
(2^n + 3^n)^(1/n)
If
Lim n to infinity of f(n) = L
and L > 0, then:
Lim n to infinity of Log[f(n)] = Log[L]
This follows from the fact that Log(x) is a continuous function. So, assuming that the limit exists and is positive we can calculate it by computing the limit of the logarithm of the function.
In this case the logarithm is:
1/n Log(2^n + 3^n) =
1/n [Log(3^n) + Log(1 + (2/3)^n)] =
Log(3) + 1/n Log(1+(2/3)^n)
For large n, the last term is:
1/n [(2/3)^n + O(2/3)^(2n)]
and it is clear that this term tends to zero for n to infinity. The logarithm of the limit is thus Log(3), therefore the limit is 3.
To make the proof rigorous, you actually have to argue this the other way around. You compute the limit if the log of the function, which is motivated by what I wrote above. Then when you find that Log[f(n)] tends to Log(3), you say:
We have a function g(n) (equal to Log of the function) which has a limit of
L = Log(3) for n to infinity. Then, because the exponential function exp(x) is continuous everywhere, we have:
Lim n to infinity of exp[g(n)] = exp(L)
This then proves the result.
wow, I'm very impressed! Thanks so much Iblis!
To find the limit as x approaches infinity of the given expression, we can use the concept of exponentials and properties of limits.
First, let's simplify the expression by factoring out the largest term within the parentheses, which in this case is 3^n:
(2^n + 3^n)^(1/n) = (3^n * (2^n/3^n + 1))^(1/n)
Now, we can rewrite the expression as:
(3^n * ((2/3)^n + 1))^(1/n)
As x approaches infinity, we can see that the term (2/3)^n becomes very small and tends towards zero, as the base (2/3) is less than 1. Additionally, the term 1 remains constant.
Therefore, as n approaches infinity, the expression simplifies to:
(3^n * (0 + 1))^(1/n) = 3^(n/n) = 3^1 = 3
Hence, the limit as x approaches infinity of (2^n + 3^n)^(1/n) is equal to 3.