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

[WK95]

Homework Statement

Prove the following:
lim x² - x = 0
x→1

Homework Equations

If 0<|x-a|<δ then |f(x)-L|<ε

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

 

[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.

 

[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.

 

[WK95]

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.

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

 

[WK95]

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.

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

 

[STEMucator]

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 $δ$, let's 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 $δ$?

 

[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
$0 < |x - 1| < δ$
$|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε$

 

[Seydlitz]

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

 

[STEMucator]

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
$0 < |x - 1| < δ$
$|(x^2 - x) - 0| < 2|x - 1| < 2(ε/2) = ε$

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.

 

[dirk_mec1]

Yes it is true if delta<1.