prove limit by definition

prove that by epsilon delta defintion. we need to prove that their is so for every their is x and so we are free to choose our epsilon and our x i choose x cannot be equal to -1 because of the denominator. and we need to choose x for which -> -> -> -> but x=-1 there so we have a problem. where is the problem in my way??? the prof solved it by cant understand why he choose such x

Re: prove limit by definition

Since the actual limit is 2/3, you will always be able to show that for x close enough to 2 that |x/(x+1)-1|>0.3, so you need to choose epsilon greater than zero but less than 0.3 (well actually any thing less than 1/3 will do). Then show that for all delta>0 there is always some x such that |x-1|<delta and |x/(x+1) - 1|>epsilon.

CB

Re: prove limit by definition

what is the problem of my way.

??

Re: prove limit by definition

Consider <_1. Then if x satisfy's 0< |x-2|< , F(x)<3/4. Hence |F(x)-1|>1/4. Choosing <_1/4 completes the proof.

Re: prove limit by definition

i cant because its for every delta

and can you tell me where am i wrong

?

Re: prove limit by definition

For any delta>1, x values in (1,3) will still have to satisfy |F(x)-1|< epsilon.

Re: prove limit by definition

It's not true that you are totally free to choose any ε that you like.

As **CaptainBlack** told you, the actual limit is 2/3.

|1 - 2/3| = 1/3. So, if you use an ε ≥ 1/3, it **is** possible to find a δ > 0 such that for all x for which 0 < | x - 2 | < δ it follows that | x/(x+1) - 1 | < ε .

That's why **CB** said you need to choose ε < 1/3 .