What is #lim_(x->oo) (ln(x+1) - ln(x-1))/(1/x)#?

To prepare the product #x (log(x + 1) - log(x - 1))# for solution by l'Hôpital's rule (where #log# here is base #e#), rewrite it as:

#lim_(x->∞) (log(x + 1) - log(x - 1))/(1/x)#

Applying l'Hôpital's rule, we get that

#lim_(x->∞) ( d/( dx)(log(x + 1) - log(x - 1)))/( d/( dx)(1/x))#

#=lim_(x->∞) (1/(x + 1) - 1/(x - 1))/(-1/x^2)#

Multiplying the numerator and denominator by #(x+1)(x-1)x^2#

#= lim_(x->∞) (2 x^2)/((x - 1) (x + 1))#

#=2 lim_(x->∞) x^2/((x - 1) (x + 1))#

Let #u = x^2#. Then #lim_(x->∞) x^2/((x - 1) (x + 1)) = lim_(u->∞) u/(u - 1)#.

So, since #(x-1)(x+1)=(x^2-1)#, we have that

#= 2 lim_(u->∞) u/(u - 1)#

Divide the numerator and denominator by #u# gives:

#=2 lim_(u->∞) 1/(1 - 1/u)#

The expression #-1/u -> 0# as #u->∞#:

#2 lim_(u->∞) 1/(1 - 1/u)=2(1)=2#

Making #(x+1)/(x-1)=1+y# and solving for #x# we have #x = (1+2/y)# then

#((x+1)/(x-1))^x equiv (1+y)^(1/y+1) = (1+y)(1+y)^(2/y)#

and #lim_(x->oo) equiv lim_(y->0)# so finally

#lim_(x->oo) x(log(x+1)-log(x-1)) = log(lim_(x->oo)((x+1)/(x-1))^x) = log(lim_(y->0) (1+y)(1+y)^(2/y)) = log(e^2) = 2#