Trying to "substitute", gives us the indeterminate form #infty - infty#.
Recall that #x = lne^x# and #lna - lnb = ln(a/b)#:
#lim_(x->infty) (x - lnx) = lim_(x->infty) (lne^x - lnx)#
#=lim_(x->infty) ln(e^x/x)#. Let #u = e^x/x#.
Now we will take the limit of #u# as #x# approaches infinity:
#lim_(x->infty) e^x/x#. Since we get the indeterminate form #infty/infty#, and #e^x# and #x#
are differentiable everywhere, we can use L'Hospital's rule:
#lim_(x->infty) (e^x/x) = lim_(x->infty) ((e^x)') / ((x)')#
#=lim_(x->infty) (e^x) = infty#.
Now we know that as #x -> infty#, #u -> infty#, and
#lim_(u->infty) (lnu) = infty#.
Therefore,
#lim_(u->infty) (lnu) = lim_(x->infty) ln(e^x/x)#
#=lim_(x->infty) (lne^x - lnx) = lim_(x->infty) (x - lnx)#.
