# Thread: delta epsilon proof for limit of multivariable function

1. ## delta epsilon proof for limit of multivariable function

lim x^2 / (x+y)
(x,y) ~> (1,2)

I find that the limit is just 1/3. but i don't know how to prove this using the delta epsilon definition. i have that |x^2 / (x+y) - (1/3)| < epsilon and sqrt((x-1)^2 + (y-2)^2) < delta. i have fiddled with the inequalities for a while but it doesn't seem to be getting anywhere. i am trying to see if i can find a relationship between epsilon and delta but am having trouble doing so. can anyone provide me a some hints in the correct direction? thanks

2. Note, if $\sqrt{(x-1)^2 + (y-2)^2} < \delta$, then you also know that $|x - 1| < \delta$ and $|y - 2| < \delta$.

3. so i combined the fractions and got (3x^2 - |x+y|)/(3|x+y|). i found that 3 </= |x+y| </= 2δ + 3 since 0<|x-1|<δ and similarly with |y-2|. so (3x^2 - |x+y|)/(3|x+y|) </= (3x^2 + 2δ + 3)/(9). then i found that |x|-1 </= |x-1| < δ so |x|< δ + 1 and x^2 < (δ+1)^2 so (3x^2 + 2δ + 3)/(9) </= (3(δ+1)^2 + 2δ + 3)/(9). so finish the proof i just have to say that given any epsilon i can pick a delta since ε = (3(δ+1)^2 + 2δ + 3)/(9).

does that look correct? i hope i didn't make any big mistakes.

4. Originally Posted by oblixps
so i combined the fractions and got (3x^2 - |x+y|)/(3|x+y|). i found that 3 </= |x+y| </= 2δ + 3 since 0<|x-1|<δ and similarly with |y-2|. so (3x^2 - |x+y|)/(3|x+y|) </= (3x^2 + 2δ + 3)/(9). then i found that |x|-1 </= |x-1| < δ so |x|< δ + 1 and x^2 < (δ+1)^2 so (3x^2 + 2δ + 3)/(9) </= (3(δ+1)^2 + 2δ + 3)/(9). so finish the proof i just have to say that given any epsilon i can pick a delta since ε = (3(δ+1)^2 + 2δ + 3)/(9).

does that look correct? i hope i didn't make any big mistakes.

|3x^2-(x+y)| not always equal to 3x^2-|(x+y)|

I think you should pick |x+y| as small as possible.

x-y>3-2delta

I found epsilon in form:

e={3d^2+8d}/{9-6d}

5. i got e={3d^2+8d + 6}/{9-6d}. because i had |3x^2 - x - y| </= 3x^2 + |x| + |y| </= 3(d+1)^2 + 2d + 3 = 3d^2+8d + 6. i got 9-6d by using the fact that |x+y|>/= 3 - 2d.