proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

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?

Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

From $\displaystyle y=x^x$, we get $\displaystyle ln(ln(y))=ln(x\cdot ln(x))=ln(x)+ln(ln(x))$. 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.

Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

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!! :D

Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

what about the g(x) the inverse of y = x^x ?

Re: proof of limy->inf,g(y)ln(ln(y))/ln(y) = 1

Quote:

Originally Posted by

**MINOANMAN** what about the g(x) the inverse of y = x^x ?

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?