Proving Limits with a Delta/Epsilon Defintion

[James98765]

I have been trying to study Calculus on my own with a text book so I don't have any kind of instructor and was wondering if I could have somebody check one of my limit proofs. I am proofing the limit with the delta/epsilon definition. I will use & to represent delta and e to represent epsilon. (<->) means equals and (->) means implied. Here is the problem:

Proove that the limit (x[sup:fjsduqje]2[/sup:fjsduqje]-1)/(X+1) = -2 as X approaches -1

Given any e>0, however small, there exists a &>0 such that:
if 0 <|x-a|<& then |f(x)-L|<e

if 0<|x + 2|<& then |f(x)+2|<e
(<->) if 0<|x+1|<& then |((x[sup:fjsduqje]2[/sup:fjsduqje]-1)/(X+1))+2|<e
(<->) if 0<|x+1|<& then |((x-1)(x+1)/(x+1))+2|<e
(<->) if 0<|x+1|<& then |(x-1)+2|<e
(<->) if 0<|x+1|<& then |x+1|<e
&=e
0<|x+1|<&
(->)|x+1|<e (because &=e)

The part that I am worried about being inaccurate is that &=e exactly. I am not sure if this still fits the definition of the limit because delta and epsilon are the same. In my thought process it is possible that delta could be the same as epsilon and you could still have an accurate limit but I was just trying to confirm with you guys before I went further. If you have any time to look this over and tell me what you think it would be great. Thanks!
-James

P.S. Sorry if the proof didn't follow a proper format. I was just trying to follow my book!

 

[tkhunny]

Proove that the limit (x[sup:2iz7luxi]2[/sup:2iz7luxi]-1)/(X+1) = -2 as X approaches -1

if 0<|x + 2|<& then |f(x)+2|<e

&=e

(->)|x+1|<e (because &=e)

1) How did that 2 sneak in there? "if 0<|x + 2|<& then"

2) There is no fixed expectation for \(\displaystyle \delta\) and \(\displaystyle \epsilon\). Equality is fine. The idea is that there is a finite limit. No one cares what it is. It is a question of existence. (Note: There is a slightly different version for infinite limits)

3) What does "because" mean?

Excellent work. Have you worked through any problems on your own? This is a great condeptual leap.

 

[James98765]

Thanks for the input
1) The 2 was in there because I was using the delta/epsilon definition. Delta is defined as the distance X is from the Limit value of x. (|x-a|) In the example problem the limit had x approaching -2 so to find the delta I said |x-(-2)|=|x+2|. That is how the 2 got in there.
2) Thanks for the delta/epsilon explanantion because I think I understand it a little bit better
3) I wrote "because" to further show that &=e because their values are the same as shown by the proof. I am not sure if that is standard format.

You asked me if I did any proofs on my own. The proof you commented on I did do on my own and my text book doesn't provide the answer to the problem so I figured I would post it to have some of you guys check my work. Thanks alot for all of your time!
-James

 

[tkhunny]

1) The 2 was in there because I was using the delta/epsilon definition. Delta is defined as the distance X is from the Limit value of x. (|x-a|) In the example problem the limit had x approaching -2 so to find the delta I said |x-(-2)|=|x+2|. That is how the 2 got in there.

I think you missed the point of my question. Try anwering again. Is it approaching -1 or -2? You seem to be switching from time to time?