Given an Epsilon, Delta proof for the following statement:
limit as x approaches 2 for f(x) = |x-2| =0
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 $\displaystyle \epsilon > 0$. Then you say, $\displaystyle |x-2| < \epsilon$. 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.