# Homework Help: Prove a limit exists using formal definition

1. Oct 1, 2014

### Luscinia

1. The problem statement, all variables and given/known data
Calculate the value of the limit and justify your answer with the ε-δ definition of the limit.
lim (x->1) x2

2. Relevant equations
My professor gave us the hint that we have to take δ as 0<δ≤ k0 so that δ(ε)=min{k0,ε/ (k0+2)}

I'm guessing that k0 is meant to be any number though it's usually 1?

3. The attempt at a solution
I'm trying to relate |f(x)-1|<ε to |x-a|<δ
|x2-1| < ε
-ε < x2-1 <ε
-ε+1 < x2 <ε+1
(-ε+1)1/2-1 < x-1 < (ε+1)1/2-1

I'm not sure what to do from here. I'm guessing that
δ=(-ε+1)1/2-1 or (ε+1)1/2-1
but I have no clue where to go from there.

I've tried doing this another way as well to make use of a k0 from my professor's hint

|x2-1| < ε
-ε < x2-1 <ε
-ε+1 < x2 <ε+1
-ε+1 < x x < ε+1 where if we assume that |x|<k0=1, we can then assume that -ε+1 < k0x < ε+1
(-ε+1)/k0 - 1 < x-1< (ε+1)/k0 - 1
This still doesn't give me what my professor hinted at though. (I don't know what to do after that.)
Also, what does the min{_____,_______} mean?

2. Oct 1, 2014

### Staff: Mentor

k0 might be 1. It's just some unspecified number.

min{..., ...} means the minimum of the numbers in the list inside the braces.

3. Oct 2, 2014

### Luscinia

I think I managed to connect the dots together. I shouldn't have gotten rid of the -1 in |x2-1|. Instead, I should have just turned it into (x-1)(x+1) and move on from there.

Assuming that x+1<1 and δ<1,
-1<x-1<1 so if we add +2 to both sides, we get 1<x+1<3
|x+1||x-1|<3|x-1|<ε
|x-1|<ε/3

But if I assume that δ<1, then δ=min{1,ε/3} won't make sense?

4. Oct 2, 2014

### Staff: Mentor

Just assume that δ < 1. That makes |x - 1| < 1, which gives you bounds on |x + 1|, as you show below.
Now, choose δ = min{1, ε/3}. Typically, someone (else) would pick a small value for ε, so δ will typically be much smaller than 1.