# Evaluation of limit

longrob

## Homework Statement

Evaluate the limit of the following as x approaches infinity

$$\frac{e^{x}-1}{1-2e^{x}+2e^{2x}}$$

## The Attempt at a Solution

$$\frac{e^{-x}-e^{-2x}}{e^{-2x}-2e^{-x}+2}$$
which gives 0/2=0 as x approaches infinity, but apparently this is wrong.

Homework Helper
Why do you think it's wrong? I think it's right.

longrob
Because I am told the answer is $$\frac{1}{2}e^{-x}$$

Homework Helper
The limit of (1/2)e^(-x) as x->infinity is 0. Are they actually asking for the asymptotic behavior or the limiting behavior? Not the limit?

longrob
The question doesn't explicitly ask for the limit, I just assumed that's what I had to do. This is an applied maths course, not analysis. The question is asking for the long term behaviour of the solution to a differential equation.........

longrob
in case I'm not being clear, the expression I posted initially is the general solution to a DE problem that I obtained (so it might be wrong !) and the question is to show that the long term behaviour as x->infinitty is that x approaches (1/2)e^(-x)

Homework Helper
That would make sense. But I'm not sure what they mean by show. Casually you would just pick the dominant term in the numerator (e^x) (dominant meaning the ratio of any other term in the numerator and that one goes to zero) and the dominant term in the denominator (2e^(2x) and divide them. If they want you be more formal they might want you to show limit f(x)/g(x) goes to 1, where f(x) is your original expression and g(x)=(1/2)e^(-x).

longrob
Thank you very much.

longrob
Dick, Your final remark about the more formal approach requires me to know the answer, ie g(x) already, which was obtained by the "casual" approach. So how do you find g(x) more formally than using the casual approach ? Thanks a lot !