# Thread: Analysis epsilon delta limit

1. ## Analysis epsilon delta limit

I'm having trouble knowing how to rearrange the inequality
The question is

Using epsilon, delta methods show that

lim f(x)= 2+x/x^2 = 3
{x->1}

I've gotten as far as saying | 2+x/x^2 - 3 | < epsilon

so is the delta part.. 0 < x-1 < delta ?????

I know all the steps but can't reach an inequality i'm happy with... I don't want to continue if i'm doing it wrong, time is of the essence!

Do I have to make the epsilon inequality similar to the delta one when i relate them? Or have I got it all wrong

Any help would be very much appreciated!

Thank you

2. Originally Posted by DBsHo
I'm having trouble knowing how to rearrange the inequality
The question is

Using epsilon, delta methods show that

lim f(x)= 2+x/x^2 = 3
{x->1}

I've gotten as far as saying | 2+x/x^2 - 3 | < epsilon

so is the delta part.. 0 < x-1 < delta ?????

I know all the steps but can't reach an inequality i'm happy with... I don't want to continue if i'm doing it wrong, time is of the essence!

Do I have to make the epsilon inequality similar to the delta one when i relate them? Or have I got it all wrong

Any help would be very much appreciated!

Thank you
Start by noting that $\left| \frac{2 + x}{x^2} - 3\right| = \left| \frac{(3x + 2)(x - 1)}{x^2}\right|$.

And you might want to read this: http://www.mathhelpforum.com/math-he...ta-proofs.html