# Is this the right way to go about proving this?

1. Sep 13, 2006

### end3r7

I'm supposed to use the definition of limit to show that $$\lim_{\substack{x\rightarrow 3\\y\rightarrow -1}} f(x,y) = 5$$

where $$f(x,y) = x - 2y$$

I split the function into two: $$H(x,y) = x$$ and $$G(x,y) = -2y$$
It can be shown that H + G = f and that the addition of their limits equals the limit of f

So I let $$\epsilon > 0$$ and set $$\delta = \epsilon/2$$, so that whenever $$\sqrt{(x-3)^2+(y+1)^2} < \delta$$ then $$|H(x,y) - 3|= |x - 3| < \epsilon/2$$ and $$|G(x,y) - 2|= |-2y - 2|= 2 |y + 1| < \epsilon/2$$
$$|H(x,y) + G(x,y) - (3 + 2)| = |H(x,y) -3 + G(x,y) -2| = |(x-3) + 2(y+1)| <= |x-3| + |2 (y+1)| < 2 \sqrt{(x-3)^2+(y+1)^2} = 2 \delta = \epsilon$$

So, what all is wrong?

Last edited: Sep 14, 2006
2. Sep 13, 2006

### MathematicalPhysicist

i think that your mistake is when you write $$\sqrt{[(x-3)^2+(y+1)^2]}<\delta$$
i think it should be $$|x-3|<\delta_1$$
and $$|y+1|<\delta_2$$

Last edited: Sep 13, 2006
3. Sep 13, 2006

### end3r7

Why is that?
If my proof is wrong, can anybody explain how to make it right, please. I've never had to prove anything calc related before =/

4. Sep 13, 2006

### HallsofIvy

Staff Emeritus
Well, that depends on exactly what your definition of limit (in R2) is!

The standard definition uses $\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2}$ to measure the distance between points so $\sqrt{(x-3)^2+ (y+1)^2}< \delta$ would be correct. It is possible to define "distance between points" by max($|x_1-x_0|, |y_1-y_0|$) so that loop quantum gravity's suggestion is possible but that is not the standard definition.

5. Sep 13, 2006

### end3r7

Thanks for the replies. =)

So is my proof correct?
And if I used his way, woudl I have to Set $$\delta = min(\delta_1, \delta_2) = \epsilon/2$$?

6. Sep 13, 2006

### end3r7

SOrry, I don't want to seem pushy, but I just want to know two things:
1) Am I correct?
2) Why or why not?

7. Sep 14, 2006

### end3r7

Oh crap, I forgot to write $$\delta = \epsilon/2$$