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?
Best Answer
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 $(1,1)$.