Is this the right way to go about proving this?

  • Thread starter end3r7
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  • #1
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I'm supposed to use the definition of limit to show that [tex]\lim_{\substack{x\rightarrow 3\\y\rightarrow -1}} f(x,y) = 5[/tex]

where [tex]f(x,y) = x - 2y[/tex]

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

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

So, what all is wrong?
 
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Answers and Replies

  • #2
MathematicalPhysicist
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i think that your mistake is when you write [tex]\sqrt{[(x-3)^2+(y+1)^2]}<\delta[/tex]
i think it should be [tex]|x-3|<\delta_1[/tex]
and [tex]|y+1|<\delta_2[/tex]
 
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  • #3
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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
HallsofIvy
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Well, that depends on exactly what your definition of limit (in R2) is!

The standard definition uses [itex]\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2}[/itex] to measure the distance between points so [itex]\sqrt{(x-3)^2+ (y+1)^2}< \delta[/itex] would be correct. It is possible to define "distance between points" by max([itex]|x_1-x_0|, |y_1-y_0|[/itex]) so that loop quantum gravity's suggestion is possible but that is not the standard definition.
 
  • #5
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Thanks for the replies. =)

So is my proof correct?
And if I used his way, woudl I have to Set [tex]\delta = min(\delta_1, \delta_2) = \epsilon/2[/tex]?
 
  • #6
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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
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Oh crap, I forgot to write [tex]\delta = \epsilon/2[/tex]
 

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