limit help (differrent question)?

Question

hoiw do you find the limit of n^(1/n)

Answer 1

[n→∞]lim n^(1/n)

In cases where exponential functions are involved, it is frequently easier to find the logarithm of the limit than it is to find the limit directly. So take the log of this:

ln [n→∞]lim n^(1/n)

Since the logarithm is continuous, we may move it inside the limit:

[n→∞]lim ln n^(1/n)

Using the laws of logarithms:

[n→∞]lim 1/n ln n

Or rearranging:

[n→∞]lim ln n/n

This is now a limit of the indeterminate form ∞/∞, however if you know about the limiting behavior of logarithmic and exponential functions, then you know that the denominator grows much more quickly than the numerator, so the limit is 0. If you don't, however, L'hopital's rule may be employed to produce:

[n→∞]lim (1/n)/1

Which is obviously zero. Thus we have:

ln [n→∞]lim n^(1/n) = 0

But we are after [n→∞]lim n^(1/n), not ln [n→∞]lim n^(1/n). Therefore, we must exponentiate both sides, thus obtaining:

[n→∞]lim n^(1/n) = e^0 = 1.

And now we are done.

Answer 2

Plug some values into your calculator and see where the values lead. This should give you a good idea of the limit. I'll call it x.

To prove your suspicion, try the sandwich method. Find a function f(n) with the same limit (x) such that you can easily prove the limit f, and f(n) > n^(1/n) for sufficiently large n (which you'll have to specify). Since f(n) > n^(1/n) > x, as f(n) approaches x, so must n^(1/n).

Answer 3

1^(1/1) = 1 lowest
2^(1/2) = 1.414213
3^(1/3) = 1.442249 highest
4^(1/4) =1.414213

N^(1/N)'s limits are 1 and 1.442249
Assuming N is a whole number