# Prove the limit is true using the epsilon,delta definition of a limit.

1. Aug 1, 2013

### WK95

1. The problem statement, all variables and given/known data
Prove the following:
lim x2 - x = 0
x→1

2. Relevant equations

If 0<|x-a|<δ then |f(x)-L|<ε

3. The attempt at a solution
Part I - Set up
0<|x-a|<δ |f(x)-L|<ε
If 0<|x-1|<δ then |(x^2 - x) - 0|<ε
x|x-1|<ε
x|x-1|<ε
|x-1|<ε/x
δ=ε/x

Part II - Proof
Given ε < 0, choose δ=ε/x
If 0<|x-1|<δ then
|(x^2 - x) - 0| = |x^2 - x| = x|x - 1| < xδ = x(ε/x) = ε

2. Aug 1, 2013

### dirk_mec1

No! Delta may not be dependent on x. Also your definition is incorrect, it is for every ε>0 there is a ... etc.

Try this:

|x||x-1| <δ|x| = δ |x-1+1|<δ^2+δ < δ if δ < 1,..., etc.

3. Aug 1, 2013

### Seydlitz

I'm afraid you can't set $δ=ε/x$. It needs to be a particular number not an arbitrary value. Try setting up $|x||x-1|$ and then apply triangle inequality to find the upper bound $f(x)$ can take. It can be assured that |x-1| will at least be smaller than 1.

4. Aug 1, 2013

### WK95

I'm not quite sure why you added 1 to |x - 1|
|x||x-1| < δ|x| = δ|x-1+1| < δ^2 + δ < δ
|x^2 - x| < δ|x-1+1| < δ^2 + δ < δ <== I'm not getting this part

5. Aug 1, 2013

### WK95

How can it be assured that |x - 1| will be smaller than 1? In a formal proof, how would I show this?

I found the Triangle Inequality to be |A + B| ≤ |A| + |B|

Set up
0<|x - a|<δ |f(x) - L|<ε
If 0<|x - 1|<δ then |(x^2 - x) - 0|<ε
x|x-1|<ε

Last edited: Aug 1, 2013
6. Aug 1, 2013

### Zondrina

So you have this :

$|x^2 - x| = |x||x-1| < δ|x|$

Now this is quite a common trick and you should use it often. It allows you to use the triangle inequality on your second factor :

$|x| = |x-1+1| ≤ |x-1| + 1 < δ + 1$

Therefore :

$|x^2 - x| = |x||x-1| < δ|x| < δ(δ+1)$

Now at this point, it's very useful to place a bound on $δ$, lets say $δ ≤ 1$ to make life simple. So we get :

$δ(δ+1) ≤ 2δ$

Now you can find $δ$ in terms of $ε$.

What is your final value(s) for $δ$?

7. Aug 1, 2013

### WK95

Here is what I have now.
$|(x^2 - x) - 0| = |x - 1| |x|$
$= |x - 1| |(x - 1) + 1| ≤ |x - 1| (|x - 1| + |1|)$
$= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|$
assuming
$|x - 1| < 1$
Given
$ε > 0, δ = min{1, ε/2}$
Then
$0 < |x - 1| < δ$
$|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε$

Last edited: Aug 1, 2013
8. Aug 2, 2013

### Seydlitz

It looks fine to me. $δ=min(1,\frac{ε}{2})$

9. Aug 2, 2013

### Zondrina

I'm not so sure I agree with the third line :

$= |x - 1|^2 + |x - 1| < |x - 1| + |x - 1| = 2|x - 1|$

How were you able to deduce $|x-1|^2 < |x-1|$. This is certainly not true in general. Your value for $δ$ is indeed correct though.

10. Aug 2, 2013

### dirk_mec1

Yes it is true if delta<1.