Prove a limit exists using formal definition

[Luscinia]

Homework Statement

Calculate the value of the limit and justify your answer with the ε-δ definition of the limit.
lim (x->1) x²

Homework 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?

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?

 

[Mark44]

Homework Statement

Calculate the value of the limit and justify your answer with the ε-δ definition of the limit.
lim (x->1) x²

Homework 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?

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?

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

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

 

[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?

 

[Mark44]

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,

Just assume that δ < 1. That makes |x - 1| < 1, which gives you bounds on |x + 1|, as you show below.

-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?

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.