We start with quite a common trick when dealing with variable exponents. We can take the natural log of something and then raise it as the exponent of the exponential function without changing its value as these are inverse operations - but it allows us to use the rules of logs in a beneficial way.
#lim_(xrarroo) (ln(x))^(1/x) = lim_(xrarroo) exp(ln((ln(x))^(1/x)))#
Using the exponent rule of logs:
#=lim_(xrarroo) exp(1/xln(ln(x)))#
Notice that it is the exponent that varies as #xrarroo# so we can focus on it and move the exponential function outside:
#=exp(lim_(xrarroo)(ln(ln(x))/x))#
If you look at the behaviour of the natural log function you will notice that as x tends to infinity, the value of the function also tends to infinity, albeit very slowly. When we take #ln(ln(x))# we have a variable inside the log function that tends to infinity very slowly, meaning that we have an overall function that tends to infinity EXTREMELY slowly. The graph below only ranges up to #x=1000# but it demonstrates the extremely slow growth of #ln(ln(x))# even in comparison to the slow growth of #ln(x)#.
From this behaviour, we can infer that #x# will exhibit much faster asymptotic growth and that the limit of the exponent will therefore be zero. #color(blue)("This means that overall limit = 1.")#
We can also tackle this point with L'hopital's rule. We need the limit to be in indeterminate form, ie #0/0 or oo/oo# so we check that this is the case:
#lim_(xrarroo)ln(ln(x)) = ln(ln(oo)) = ln(oo) = oo#
#lim_(xrarroo) x = oo#
This is indeed the case so limit becomes:
#=exp(lim_(xrarroo)((d/(dx)(ln(ln(x))))/(d/(dx)x)))#
To differentiate #y = ln(ln(x))# recognise we have #y(u(x))# and use the chain rule
#(dy)/(dx) = (dy)/(du)(du)/(dx)#
#u = ln(x) implies (du)/(dx) = 1/x#
#y = ln(u) implies (dy)/(du) = 1/u = 1/(ln(x))#
#therefore (dy)/(dx) = 1/(ln(x))*1/x = 1/(xln(x))#
Derivative of #x# is #1#. Limit becomes:
#=exp(lim_(xrarroo)((1/(xln(x)))/1)) = exp(lim_(xrarroo)(1/(xln(x))))#
We have addressed that both functions on the denominator tend to infinity so we have
#exp(1/oo) = exp(0) = 1#
