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My textbook asks the question

$$f(x,y) = \frac{x-y}{x^2-y^2}$$ Does $f(x,y)$ have a limit as $(x,y) \rightarrow (1,1)$?

First step of finding a value is easy if we approach $(1,1)$ on the y-axis:

$$\lim_{(x,0) \rightarrow (1,1)} f(x,y) = \lim_{(x,0) \rightarrow (1,1)} \frac{x-0}{x^2 - 0} = 1 $$

So if the limit exists it should be 1.

Next is to examine it using the formal ($\epsilon$-$\delta$) definition:

Choose $\delta,\epsilon > 0$ where $\delta = \delta(\epsilon)$ such that for all $x,y \in D_f$, $|f(x,y)-L| < \epsilon$ whenever $0 < \sqrt{(x-a)^2 + (y - b)^2} < \delta$

In this case the inequalities become

$$ \left| \frac{x-y}{x^2-y^2} - 1\right| < \epsilon \quad \text{whenever} \quad 0 < \sqrt{(x-1)^2 + (y-1)^2} < \delta$$

Using the identity $(a+b)(a-b) = a^2 + b^2$ on the left inequality, and expanding the right inequality, we get

$$ \left| \frac{1}{x+y} - 1\right| < \epsilon \quad \text{whenever} \quad 0 < \sqrt{x^2 + y^2 - 2x - 2y + 2} < \delta$$

But then I'm stuck, I can't get them to look like eachother, so I can neither prove nor disprove that a limit exists. What do I do from here?

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  • $\begingroup$ You're wrong you cant approach it from the y axis as then y goes to 0 not to 1. and the limit is $2$ not $1$. Also welcome to the site, sorry if my comment sounds rude,I've upvoted your question because it is a well written question. $\endgroup$ – kingW3 Jul 11 '17 at 18:36
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Just remark that in a neighborhood of $(1,1)$, $f(x,y)={{x-y}\over{x^2-y^2}}={{x-y}\over{(x-y)(x+y)}}={1\over{x+y}}$ which is defined at $(1,1)$.

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  • $\begingroup$ Uh... yeah. Now I feel stupid. $\endgroup$ – Mossmyr Jul 11 '17 at 23:06

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