Question

prove Limit(2*x/x-3)=2 x=infinity
I get to |2*x/x-3 - (2)| < epsilon and then Im stuck. can anyone help?

Answer 1

Distribute your denominator. The x in the numerator will cancel, and you should be able to solve it from there.

Answer 2

I get x/x-3<epsilon I then need to fine the interval for epsilon

Answer 3

hmmm what exactly are you trying to prove? that
\[\lim_{x\to \infty}\frac{2x}{x-3}=2\]?

Answer 4

there is no "interval for \(\epsilon\)" what you are trying to show is that given any
\(\epsilon>0\) there is an \(N\) for which if \(x>N\) then \(|\frac{2x}{x-3}-2|<\epsilon\)

Answer 5

when you subtract you do not get(\frac{x}{x-3}\) but rather you get
\[|\frac{6}{x-3}|<\epsilon\] and it should be clear how to make that small, by making \(x-3\) large as needed

Answer 6

perfect thanks!
therefore [6/epsilon +3, infinity) ...this is what I was looking for :)