From , we get . Now replace the limit as y goes to infinity, by the limit as x goes to infinity with the function written in terms of x.
Hi everyone,
Stuck on the following proof, any help would be appreciated! (Calculus a Complete Course, Adams, 3rd Ed., Ch3 Ch Review, in case anyone has the text)
I *think* I have part a) ok
a) Show that the fuction f(x) = x^x is strictly increasing on [e^-1, if[
y = x^x
ln(y) = x*ln(x)
d/dx => 1/y * y' = x * 1/x + ln(x)
y' = y(1+ln(x)) = x^x *(1+ln(x))
if x = 1/e, ln(x) = -1,
therefore y' = y (1 -1)= 0
for n > e^-1
y' = n^n * (1+ln(n)) > 0
induction, k = n0, for k+1
y' = (k+1)^(k+1)*(1+ln(k+1)) > 0
Derivative is always greater than zero so strictly increasing
I am lost on b)
b) If g is the inverse function to f of part (a), show that
lim y-> inf g(y)ln(ln(y))/ln(y) = 1
hint: start with the eq. y = x^x and take the ln of both sides twice
y = x^x
ln(ln(y)) = ln(ln(x^x))
Now what? I spent ages trying to figure this out and got no where
Cheers?
Gah! I was missing such a simple step! I totally forgot about the rule for the summation of logarithms (fail...)
I think I got it now.
y = x^x
=> ln(ln(y)) = ln(x*ln(x)) = ln(x) + ln(ln(x)), using ln(xy) = ln(x) + ln(y)
as y -> inf x -> inf since y=x^x
Therefore,
use lim x->inf, subbing in x^x
lim lim x->inf, g(y)ln(ln(y))/ln(y) = g(x^x)[ln(x)+ln(ln(x))]/(x*ln(x))
since g is the inverse of f, g(x^x) = x
Therefore,
g(x^x)[ln(x)+ln(ln(x))]/(x*ln(x)) = x*[ln(x)+ln(ln(x))]/(x*ln(x)) = [ln(x)+ln(ln(x))]/ln(x)
= ln(x)/ln(x) + ln(ln(x))/ln(x)
= 1 + ln(ln(x))/ln(x)
as x -> inf, ln(ln(x))/(ln(x)) = 0, since ln(x) increases faster than ln(ln(x)) as x->inf, [d/dx ln(x) = 1/x, d/dx ln(ln(x)) = 1/(x*ln(x))]
therefore as x = inf, 1 + ln(ln(x))/ln(x) = 1
Thanks a lot! Much love to these forums!!
Property of inverse functions: inverse_f(f(x)) = x for all x in the domain of f.
Therefore, since g is the inverse of f, g(f(x)) = g(x^x) = x
Inverse function - Wikipedia, the free encyclopedia <- "Inverses and composition"
Seems ok to me. Did I do something wrong?