# Thread: Multi-variable limit proof using delta and epsilon

1. ## Multi-variable limit proof using delta and epsilon

Hi,

I am trying to create a proof that the limit as (x,y) approaches (1,0) of x + 2y is equal to 1. I have researched online and read my textbook over and over again and understand what the epsilon/delta limit definition is. However, I am having trouble applying it to this multi-variable situation. Can anyone help? It's truly appreciated! Thanks.

2. Originally Posted by dyllusionist
Hi,

I am trying to create a proof that the limit as (x,y) approaches (1,0) of x + 2y is equal to 1. I have researched online and read my textbook over and over again and understand what the epsilon/delta limit definition is. However, I am having trouble applying it to this multi-variable situation. Can anyone help? It's truly appreciated! Thanks.
You must prove that for every $\varepsilon>0$ there exists some $\delta>0$ such that $d\left((x,y),(1,0)\right)=\sqrt{(x-1)^2+y^2}<\delta\implies \left|x+2y-1\right|<\varepsilon$

3. Thanks for that quick response! Wow.

Unfortunately, this is exactly where I am right now in the problem. I'm trying to figure out how to write my epsilon in terms of delta. My brain can't work with that square root or absolute value sign at all right now.

I now have written down that the absolute value of x - 1 is less than the absolute value of x - 1 + 2y which is less than epsilon. However, I am stuck again. Can I get a push in the right direction?

Once again, thank you for your help. This is an incredible resource.

4. ## possible limit proof

abs(x - 1) = sqrt((x-1)^2) less than = sqrt((x-1)^2 + y^2)

also,

2abs(y) = 2sqrt(y^2) less than = sqrt((x-1)^2 + y^2)

so

abs(x + 2y - 1) less than = abs(x-1) + 2abs(y) less than = 3sqrt((x-1)^2 + y^2) less than delta.

choose delta = epsilon / 3. This should work.

5. My mistake...in the second line, the final sqrt should be multiplied by the 2.

I suppose if I figure out LaTex, posts would look better?