how to compute the limit of f(x)=(log x)^(log x)?

Question

how to compute the limit of f(x)=(log x)^(log x)? the logs have base 2. Is there a way to simplify the function further?

many thanks in advance.

Answer 1

You are talking about the limit as x goes to infinity? The limit of log₂(x) is infinity, and the limit of y^y as y goes to infinity is infinity; it follows that the limit here is infinity.

Answer 2

There are three potentially interesting values which x might approach: 0, 1, and infinity. As x approaches infinity, both the base ln(x) and the exponent ln(x) grow without bound, so the function goes to infinity. As x approaches 0, serious problems occur because we take negative numbers (ln(x) is negative for 0<x<1) to strange powers -- impossible over the real numbers. The function would not be defined. As x approaches 1 (from the right), the function takes the so-called "indeterminant form" 0^0, which can be solved using L'Hopital's rule by taking logs (again!). I think you'll find log(x)^log(x) approaches 1, regardless of the base of the logarithm.