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

Question

Aurora Alvarado · Accepted Answer

1

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#

Christopher Agresta · Answer

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

Emilia Aguilar · Answer

For a third alternative, see below.

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#

William Abdalla · Answer

undefined To find the limit as x approaches infinity of (e^x + e^(-x))/(e^x - e^(-x)), we can simplify the expression by multiplying both the numerator and denominator by e^x. This gives us (e^x * e^x + e^x * e^(-x))/(e^x * e^x - e^x * e^(-x)). Simplifying further, we have (e^(2x) + 1)/(e^(2x) - 1).

As x approaches infinity, the terms involving e^(-x) become negligible compared to the terms involving e^x. Therefore, we can ignore them in the limit.

Taking the limit as x approaches infinity, we have (lim_(x to oo) e^(2x) + 1)/(lim_(x to oo) e^(2x) - 1).

Since e^(2x) grows exponentially as x approaches infinity, both the numerator and denominator tend to infinity.

Therefore, the limit as x approaches infinity of (e^x + e^(-x))/(e^x - e^(-x)) is equal to 1.