Calculus 2 exam tomorrow... need help with a sequence.?

Question

How would I go about evaluating the limit of this sequence as n tends to infinity?

(e^n + e^-n) / (e^(2n) - 1)

Answer 1

Rearrange slightly.

(e^n+1/e^n)/(e^(2n)-1)=

(e^(2n)+1)/((e^n)(e^(2n)-1)=

pull out the e^2n term

e^2n(1+1/e^2n)/(e^3n(1+1/e^2n)=

(1+1/e^2n)/(e^n*(1-1/e^2n))=
the numerator goes to 1 as n->∞
part of the denominator goes to 1 as n->∞ and the 1/e^n term goes to 0 as n->∞ so the expressing approaches 0 in the limit.

Answer 2

Rearrange the expression by dividing thru by e^(2n)

(e^n + e^-n) / (e^(2n) - 1)
= [e^(-n) + e^(-3n)] / [1 - e^(-2n)]

As n→∞ [e^(-n) + e^(-3n)] / [1 - e^(-2n)]
= (0 + 0) / (1 - 0) = 0