Question

How do you find the limit of `(e^x-1)/(e^x+1)` as `x->oo`?

Answer 1

`1`. L'Hôpital's rule.

Explanation:

The limit, if it exists, of `f(x)/g(x)` as `x to oo` is the same as the limit, if it exists, of `(f prime(x))/(g prime(x))`. Replacing `f(x)` with `e^x-1` and `g(x)` with `e^x+1` gives `1/1` immediately with no `x`.

Answer 2

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

Explanation:

If you multiply numerator and denominator by `e^(-x)` you can see that:

`(e^x-1)/(e^x+1) = (1-e^(-x))/(1+e^(-x))`

and as:

`lim_(x->oo) e^(-x)=0`

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