# A multivariable limit problem (epsilon-delta -proof)

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1. Mar 12, 2016

### TheSodesa

1. The problem statement, all variables and given/known data

Find the limit $$\lim_{(x,y)\to(2,2)}\frac{x^3-y^3}{x-y}$$

2. Relevant equations

$\epsilon - \delta$, baby:

If the limit L exists,
$$\forall \: \epsilon \: \exists \: \delta: 0 < \sqrt{(x-a)^2+(y-b)^2} < \delta \rightarrow |f(x,y)-L| < \epsilon$$

3. The attempt at a solution

By approaching the point (2,2) along the lines x=2 and y=2 I managed to figure out that the limit is likely 12. Now I just need to prove it, and I'm stuck.

Now
$$0 < \sqrt{(x-2)^2+(y-2)^2} < \delta \leftrightarrow (x-2)^2+(y-2)^2 < \delta^2$$

and using long division
$$x^3-y^3 = (x-y)(x^2+xy+y^2) \rightarrow \frac{x^3-y^3}{x-y}=(x^2+xy+y^2)$$

I tried deriving the expression for delta from epsilon like this:
$$\begin{split} |(x^2+xy+y^2)-12| &= |x^2-4x+4+y^2-4x+4+xy+4x+4y-20|\\ &\leq |x^2-4x+4+y^2-4x+4|+|xy+4x+4y-20|\\ &\leq \delta^2 + |xy+4x+4y-20| \end{split}$$

Almost there, except I'm stuck with that pesky |xy+4x+4y-20|, that I don't know how to get rid of. Am I taking the wrong approach to this entirely? Help, I'm desperate.

2. Mar 12, 2016

### Benoit

Why don't you use this equation and plug (2,2) instead of the complicated way ?

3. Mar 12, 2016

### TheSodesa

Now that you mention it...

$f(2,2)=2^2+2^2+2^2=12$

However, since I started this, I might as well finish it. Would you happen to know how to finish the proof? Is my approach going to work, or should I take a different route, still using $\epsilon - \delta$?

4. Mar 12, 2016

### Benoit

When you will plug (2,2) into that, it will cancel out. You have it!

5. Mar 12, 2016

### TheSodesa

I don't want to sound like a retard, but am I allowed to just plug values into epsilon-delta -proofs? I thought I was supposed to evaluate the terms upwards with $\leq$-signs until I arrive at an expression that consists of nothing but $\delta$s.

6. Mar 12, 2016

### Benoit

You're not retard at all, I was mistaken. Let me look at it again.

7. Mar 12, 2016

### Ray Vickson

If all you want is any $\delta>0$ that "works" (for a given $\epsilon > 0$)--rather than the best possible $\delta$ for that $\epsilon$--then you can always relax a bit and use
$$\begin{array}{rcl} |x^2 + xy + y^2 - 12| & = & |(x^2-4) + (xy - 4) + (y^2 - 4)| \\ &\leq & |x^2 - 4 | + |xy - 4| + |y^2 - 4| \end{array}$$
That is a lot easier to work with.

8. Mar 12, 2016

### TheSodesa

$$\begin{array}{rcl} |x^2 + xy + y^2 - 12| & = & |(x^2-4) + (xy - 4) + (y^2 - 4)| \\ &\leq & |x^2 - 4 | + |xy - 4| + |y^2 - 4| \\ &= & |(x-2+4)(x-2)|+|(y-2+4)(y-2)|+|xy-4|\\ &\leq & (|x-2|+4)|x-2| + (|y-2|+4)|y-2|+|xy-4|\\ &= & |x-2|^2 + 4|x-2| + |y-2|^2 + 4|y-2| + |xy-4|\\ &\leq & \delta^2 +4|x-2| + 4|y-2| + |xy-4|\\ &= & \delta^2 +4|(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})| + 4|(\sqrt{y}-\sqrt{2})(\sqrt{y}+\sqrt{2})| + |(\sqrt{x}\sqrt{y}+2)(\sqrt{x}\sqrt{y}-2)|\\ \end{array}$$

This is getting too complicated again. I'm not sure how I could evaluate the term with xy in it out of the expression. I'm not even sure I should be doing that since it might somehow cancel out the other 2 pesky terms.

9. Mar 12, 2016

### Ray Vickson

So, simplify it: find $\delta_x>0, \delta_y > 0$ that give $|x^2-4| < \epsilon/3$, $|xy - 4| < \epsilon/3$, etc., for $|x-2| < \delta_x$ and $|y-2| < \delta_y$. Again, all you need so is find $\delta$s that work, not necessarily the best possible ones.

10. Mar 13, 2016

### TheSodesa

Alright, so you're saying that I should simply assume, that since all of the three terms are each $<\epsilon/3$, they might as well be equal, in which case I could write $|xy-4|$ as either $|x^2-4|$ or $|y^2-4|$?

Therefore if $|x-2|<\delta_x$,

$$\begin{split} |x^2-4| &=|x+2||x-2| < |x+2|\delta_x < \frac{\epsilon}{3}\\ &\leftrightarrow |x+2| < \frac{\epsilon}{3\delta_x} &|\text{and also} \\ &|y+2| < \frac{\epsilon}{3\delta_y} &|\text{because the expressions have the same form} \end{split}$$

I'm not sure how this is going to help me get forward, since if I plug these expressions in, I'm just going to end up with $\epsilon = \epsilon$

11. Mar 13, 2016

### Ray Vickson

You are claiming I said things I never said! I did not say they could be assumed to be equal; I just required that all three of them be less than $\epsilon/3$.

Actually, there is nothing sacred about $\epsilon/3$: we could take $|x^2-4| < \epsilon/10$, $|xy-4| <7 \epsilon/10$ and $|y^2-4| < 2 \epsilon/10$, but choosing 1/3,1/3,1/3 seemed easiest.

Anyway, that is my last word on this topic; you can always look in a textbook or on-line for standard techniques for dealing with such proofs. Nothing I said was original; it was all from books on calculus and/or analysis.

12. Mar 15, 2016

### TheSodesa

Alright. Thanks for bearing with me. Our lecturer wasn't expecting us to actually learn how to do a proper epsilon-delta proof, just to get the general idea of what it is. This exercise didn't actually require the use of the method, although a later one kind of did, and even that wasn't a rigorous epsilon delta proof, where one constructs delta from epsilon, but one that used a geometric trick to find a suitable epsilon.

I'm going to mark this one as solved. Thanks for the help, people.

13. Mar 16, 2016

### Ray Vickson

In that case I can say more: for $x,y$ close to 2 we can certainly assume $|x|,|y| < 3$, so:
$$\begin{array}{l} |x^2-4| = |(x^2-2x)+(2x-4)| \leq |x||x-2| + 2|x-2| < 5|x-2| \\ |xy-4| = |(xy-2x)+(2x-4)| \leq |x||y-2| + 2|x-2| < 3|y-2| + 2|x-2| \\ |y^2-4| =|(y^2-2y)+(2y-4)| \leq |y| |y-2| + 2 |y-2| < 5|y-2| \end{array}$$
Thus
$$|x^2+xy+y^2-12| < 7|x-2| + 8|y-2| < \epsilon$$
if $|x-2| < \epsilon/15$ and $|y-2| < \epsilon/15$, so we can take $\delta_x = \delta_y = \delta = \epsilon/15$.

14. Mar 25, 2016

### TheSodesa

This is a bit late, but thanks. This is something I might be able to refer to later. I heard a senior maths student say that it took him 3 years to properly get his head around this business, so I guess I shouldn't be too worried. Yet.

15. Mar 25, 2016

### Ray Vickson

I'm glad it is a bit late, because PF rules forbid us from solving your problems for you before you hand them in. I was remiss in not checking first!

16. Mar 26, 2016

### TheSodesa

By "this" I was referring to my thanks to you being late, not your response. It did occur to me that using the word 'this' twice in different contexts might confuse you (or anybody, for that matter), but got lazy and didn't bother to be more elaborate in my wording.

Anyway, thanks again for your efforts.