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

3 Answers
Dec 4, 2017

1

Explanation:

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

Dec 4, 2017

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

Explanation:

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.

Dec 4, 2017

For a third alternative, see below.

Explanation:

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