# Limits, find the smallest +ve number N

1. Sep 15, 2009

### John O' Meara

Find the smallest positive number N such that for each x in the interval (N,+infinity) the value of $$f(x)= \frac{x}{x+1}$$ is within .01 units of L=1.
From the definition $$|f(x)-L|<\epsilon \mbox{ if } x > N, \epsilon>0, N>0$$. It follows that $$|\frac{x}{x+1}-1| < \epsilon \mbox{ if } x > N \mbox{ therefore } |\frac{-1}{x+1}| < \epsilon \mbox{ if } x > N$$ Now if I move the minus sign across to epsilon I get $$|\frac{1}{x+1}| > -\epsilon \mbox{ if } x > N$$. But epsilon is positive and so is x and so is N yet I will get a negative answer for N according to the last statement. Can anyone clarify the argument needed to get $$|\frac{-1}{x+1}| = |\frac{1}{x+1}|$$ so that N can remain positive? Thanks

2. Sep 15, 2009

### boboYO

i didn't read all that (have to catcha bus) but can't you just go x/(x+1) = 1-0.01 and solve for x?

3. Sep 15, 2009

### John O' Meara

Yes, you can just solve for x as you say.

4. Sep 15, 2009

### Hurkyl

Staff Emeritus
How'd you do that?

P.S. use \left| and \right| to make big bars.

5. Sep 15, 2009

### HallsofIvy

No, since that is an absolute value, it is larger that negative epsilon for all n.
$|\frac{-1}{x+1}|= |\frac{1}{x+1}|< \epsilon$ says that $-\epsilon< \frac{1}{x+1}< \epsilon$

6. Sep 15, 2009

### John O' Meara

Hurkyl, I used a numerical example -5< 4 therefore 5 > -4. Thanks for pointing out my error.Thanks all for the replies.