Question

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

Answer 1

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`

Answer 2

`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.

Answer 3

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`