# Delta Epsilon Proof

• October 19th 2009, 03:30 AM
UC151CPR
Delta Epsilon Proof
Given an Epsilon, Delta proof for the following statement:

limit as x approaches 2 for f(x) = |x-2| =0
• October 19th 2009, 03:36 AM
harbottle
This shouldn't be too hard. Just take $\epsilon = \delta$.

$|x-2|<\epsilon \Rightarrow |f(x)-0| = |x-2| < \delta$ for any $\epsilon,\delta > 0$
• October 19th 2009, 03:40 AM
Plato
Quote:

Originally Posted by UC151CPR
Given an Epsilon, Delta proof for the following statement: limit as x approaches 2 for f(x) = |x-2| =0

Here is the basic idea:
$\left| {f(x) - 0} \right| = \left| {x - 2} \right| < \delta = \varepsilon$
• October 19th 2009, 03:45 AM
UC151CPR
Would this be how to setup the proof:

Let epsilon > 0.

0 < |x-2|< delta

||x-2| - 0| < epsilon

||x-2| - 0| = ||x-2|| = |x-2|

Now, |x-2| < delta ---> |x-2| < epsilon

In order to achieve ||x-2|-0| < epsilon, delta < epsilon.

Thus chosing 0 < delta < epsilon, we have |x-2|< delta and ||x-2| - 0| < epsilon.

Hence the limit is proved.

Does that work as a delta epsilon proof. If not, can you please show me how you would set this up mathematically.
• October 19th 2009, 03:59 AM
harbottle
Quote:

Originally Posted by UC151CPR
Would this be how to setup the proof:

Let epsilon > 0.

0 < |x-2|< delta

||x-2| - 0| < epsilon

||x-2| - 0| = ||x-2|| = |x-2|

Now, |x-2| < delta ---> |x-2| < epsilon

In order to achieve ||x-2|-0| < epsilon, delta < epsilon.

Thus chosing 0 < delta < epsilon, we have |x-2|< delta and ||x-2| - 0| < epsilon.

Hence the limit is proved.

Does that work as a delta epsilon proof. If not, can you please show me how you would set this up mathematically.

You start with $\epsilon > 0$. Then you say, $|x-2| < \epsilon$. This is because we want the limit as x approaches 2.

|x-2| represents the distance between x and 2. We would like to say, for ANY value of epsilon, that is, epsilon can be as small as we like, we can make |f(x)-L| as small as we like. |f(x) - L| is the distance between the value of the function and the limit, in this case, zero.

The basic idea is that |f(x)-L| will be less than some quantity which depends only on the small quantity we chose earlier. So, the closer x becomes to 2, the closer f(x) will become to 0.

Anyway, back to the problem. we say |x - 2| < epsilon.

Now look at |f(x) - 0| = |f(x)|. But this is just |x-2|. What can we say about this quantity?

Well, it's smaller than epsilon. So, take delta equal to epsilon, and the proof is complete.
• October 19th 2009, 04:15 AM
UC151CPR
So...

Let epsilon > 0.

0 < |x-2|< delta

||x-2| - 0| < epsilon

||x-2| - 0| = ||x-2|| = |x-2|

Now, |x-2| < delta ---> |x-2| < epsilon

delta = epsilon hence proving the limit?
• October 19th 2009, 04:21 AM
harbottle
Quote:

Originally Posted by UC151CPR
So...

Let epsilon > 0.

0 < |x-2|< delta

||x-2| - 0| < epsilon

||x-2| - 0| = ||x-2|| = |x-2|

Now, |x-2| < delta ---> |x-2| < epsilon

delta = epsilon hence proving the limit?

No. Start by assuming |x - 2| < epsilon. Now, look at |f(x) -0|. Using the information |x-2| < epsilon, find a delta , which depends on epsilon and only on epsilon, such that |f(x)-0| < delta.

In this problem, it just so happens that delta = epsilon is that value.
• October 19th 2009, 04:24 AM
UC151CPR
Can you show me how I would write this proof mathematically that way I can see the necessary steps/statements that I need to make for my other proofs?
• October 19th 2009, 04:27 AM
harbottle
I did, in my first post.

Let $\epsilon > 0$, and let $\delta(\epsilon) = \epsilon$.

Then $

|x-2|<\epsilon \Rightarrow |f(x)-0| = |x-2| < \epsilon = \delta.
$