Find $\lim_{(x,y)\to(1,1)}\frac{x^y-y(x-1)-1}{(x-1)^2+(y-1)^2}$ or show that it doesn't exist.

Question

I have tried many different methods, I can't find any upper bound for squeeze theorem. Polar coordinates create big, messy limit which is hard to evaluate. WolframAlpha suggests that the limit doesn't exist, but on the graph it looks like the limit is $0$. I tried finding counterexample but all lines going through $(1,1)$ produce $0$ as the limit.

Answer 1

For $y<1$ we obtain: $$x^y=1+y(x-1)-\frac{(1-y)y(x-1)^2}{2}+\frac{(2-y)(1-y)y(x-1)^3}{6}-\frac{(3-y)(2-y)(1-y)y(\theta x)^4}{24}\leq$$ $$\leq1+y(x-1)-\frac{(1-y)y(x-1)^2}{2}+\frac{(2-y)(1-y)y(x-1)^3}{6}.$$ Thus, $$\frac{x^y-y(x-1)-1}{(x-1)^2+(y-1)^2}\leq\frac{(x-1)^2}{(x-1)^2+(y-1)^2}\left(-\frac{y(1-y)}{2}+\frac{y(1-y)(2-y)(x-1)}{6}\right)\rightarrow0.$$ Can you end it now?

Answer 2

On the line $y = x$ your function becomes

$$f(x, x) = \frac{x^x-(x-1) x-1}{2 (x-1)^2}$$

And it's rather easy to check that $\lim_{x\to 1} f(x, x) = 0$.

On the line $y = -x$ the function becomes

$$f(x, -x) = \frac{x^{-x}+(x-1) x-1}{2 \left(x^2+1\right)}$$

And again $\lim_{x\to 1} f(x, -x) = 0$

You can go even on the lines $y = \alpha x$ for $\alpha \in \mathbb{R}$ and you will get the same result. Indeed:

$$f(x, \alpha x) = \frac{x^{a x}-a (x-1) x-1}{(a x-1)^2+(x-1)^2}$$

And again $\lim_{x\to 1} f(x, \alpha x) = 0$