How do you find the limit of #(e^x-1)/(e^x+1)# as #x->oo#?

2 Answers
Dec 30, 2016

#1#. L'Hôpital's rule.

Explanation:

The limit, if it exists, of #f(x)/g(x)# as #x to oo# is the same as the limit, if it exists, of #(f prime(x))/(g prime(x))#. Replacing #f(x)# with #e^x-1# and #g(x)# with #e^x+1# gives #1/1# immediately with no #x#.

Dec 30, 2016

#lim_(x->oo) (e^x-1)/(e^x+1) = 1#

Explanation:

If you multiply numerator and denominator by #e^(-x)# you can see that:

#(e^x-1)/(e^x+1) = (1-e^(-x))/(1+e^(-x))#

and as:

#lim_(x->oo) e^(-x)=0#

#lim_(x->oo) (e^x-1)/(e^x+1) = lim_(x->oo) (1-e^(-x))/(1+e^(-x))= 1#