Is this the right way to go about proving this?

[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?

 

[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

 

[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 =/

 

[HallsofIvy]

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.

 

[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?

 

[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?

 

[end3r7]

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