How do you find the #lim_(x to oo) (e^x+e^-x)/(e^x-e^-x)#?

We can manipulate and adjust this via multiplieying both numorator and denominator by #e^x#

#((e^x + e^(-x) )/( e^x - e^(-x) ) )(e^x/e^x) #
#= (e^(2x) +1 )/(e^(2x) - 1)#

We know that as #x# gets large, #e^(2x)+1 approx e^(2x)#
and also #e^(2x) - 1 approx e^(2x) #

So hence limit becomes;

#lim_(x->oo) e^(2x) / e^(2x) #

#= lim_(x->oo) 1 #

#=1#

Given:

#lim_(x to oo) (e^x+e^-x)/(e^x-e^-x)#

Add 0 to the numerator in the form #-e^-x+e^-x#

#lim_(x to oo) (e^x-e^-x+e^-x+e^-x)/(e^x-e^-x)#

Combine like terms:

#lim_(x to oo) (e^x-e^-x+2e^-x)/(e^x-e^-x)#

Separate into two fractions:

#lim_(x to oo) (e^x-e^-x)/(e^x-e^-x)+(2e^-x)/(e^x-e^-x)#

The first fraction becomes 1:

#1 + lim_(x to oo) (2e^-x)/(e^x-e^-x)#

Multiply the fraction by 1 in the form of #e^x/e^x#

#1 + lim_(x to oo) e^x/e^x(2e^-x)/(e^x-e^-x)#

Perform the multiplication:

#1 + lim_(x to oo) 2/(e^(2x)-1)#

The limit becomes 0; leaving only the 1.

Multiply numerator and denominator by #e^-x#

#((e^x + e^(-x) )/( e^x - e^(-x) ) )(e^-x/e^-x) #
#= (1+e^(-2x))/(1-e^(-2x))#

We know that as #x# increases without bound, #e^x# also increases without bound, so

#e^(-2x) = 1/e^(2x)# goes to #0#

So the limit becomes;

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