An epsilon-delta proof of a limit

Problem:

Give an epsilon-delta proof for the existence of the limit,

> ;

```                                              2    2
x  - y
lim         -------
(x, y) -> (0, 0)   x + y```

Solution:

> f := (x,y) -> (x^2-y^2)/(x+y):
 First notice that if the limit exists it must be the value obtained when we compute it along any path aproaching (0,0). So let us take the simple path y=x as x->0. We have,

> ;

```                                       2    2
x  - x
lim    ------- = 0
x -> 0   x + x```
 We conclude that the only candidate for the limit is 0. For the limit to be zero we must show that, for any epsilon > 0 there is delta > 0 such that when 0 < |(x,y)-(0,0)| < delta we have |f(x,y)-0| < epsilon

 in order to show this we factorize the numerator of f(x,y) as,

> factor(x^2-y^2);

`                                (x - y) (x + y)`
 and then for all (x,y) different from (0,0) we have

> 'f(x,y)' = simplify(f(x,y));

`                                f(x, y) = x - y`
 At this point we can be certain that the limit exists and that it is in fact 0. We could see it, that is, if we accept the fact that (x-y) is a continuous function at (0,0). But to show that (x-y) is continuous you need to show the existence of the limit so the complete argument is circular. There is no way but to use the epsilon-delta proof. One more point: If we use the theorem (limit of a sum) = (sum of the limits) the epsilon-delta proof would be much easier... but why making it easy if we can make it more complicated? By the way the real practical value of the epsilon-delta definition is that it allows to show theorems like the one above. OK back to the proof. We need to show that, |f(x,y)-0| < u(delta) = (some function of delta) and then find a delta such that u(delta) < epsilon if u() is increasing with inverse v() then v() is also increasing and by applying v on both sides of the inequality we obtain, delta < v(epsilon) and we are DONE! That's the usual plan for epsilon-delta proofs. That's at least the theory. Now back to the dirty specifics of the real (mathematical) world

> ;

```                                        2    2  1/2
|f(x,y)-0| = |x-y| < |x|+|y| < (2 (x  + y ))   = sqrt(2) delta```
 Ok the last inequality needs a justification. Here it is:

Lemma:

 For all (x,y) we have,

> ;

```                2    2  1/2
|x|+|y| < (2 (x  + y ))
```

Proof:

 here is the trick,

> ;

```                           2        2                        2
0 <  (| x | - | y |)  = | x |  - 2 | x | | y | + | y |```
 from where we deduce that

> ;

```                                           2        2
2 | x y | < | x |  + | y |```
 Put this on the ONE hand. On the OTHER hand put

> ;

```                             2        2                    2
(| x | + | y |)  = | x |  + 2 | x y | + | y |```
 and use the stuff you got on the ONE hand to show that

> ;

```                                   2           2        2
(| x | + | y |)  < 2 (| x |  + | y | )```
 Finally use the fact that sqrt() is an increasing function and take sqrt() on both sides of this last inequality and we are done with the Lemma!

 With the Lemma under the belt we do the final epsilon-delta hunky punky (i.e. invert the u to get the v... remember?) and get:

> ;

```                           epsilon
delta < -------
sqrt(2)```

Yippiiii....

Mom, I know how to do epsilon-delta proofs!!!

Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>