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.
Limit as x approaches infinity of
(2^n + 3^n)^(1/n)
3 answers
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.
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!
5 answers