Lim (x->+oo) (ln(1+ax))/(ln(1+bx))=?

I forget to mention that a<b
Sure it is zero but I do not know how to formally proof it.

1 Answer
Dec 30, 2017

#lim_(x->+oo) (ln(1+ax))/(ln(1+bx)) =1#

Explanation:

#lim_(x->+oo) (ln(1+ax))/(ln(1+bx)) #

Assume #{a,b}>0#

Here we have a limit of indeterminate form #oo/oo#.

Hence we can apply L'Hopital's rule:

#lim_(x->+oo) (ln(1+ax))/(ln(1+bx)) = lim_(x->+oo) (d/dxln(1+ax))/(d/dxln(1+bx)#

#=lim_(x->+oo) (1/(1+ax)*a)/(1/(1+bx)*b)#

#= a/b * lim_(x->+oo) (1+bx)/(1+ax)#

#= a/b * lim_(x->+oo) (1/x +b)/(1/x +a)#

#= a/b *(0+b)/(0+a)#

#= a/b * b/a =1#