What is $\lim_{(x,y)\to(0,0)} \frac{\sin(x-y)}{x^3-y^3}$?

Question

I've tried to prove it doesn't exist then I tried to find some sort of limited function inside of it that would help me out, nothing worked, I ran out of ideas for now.

Answer 1

Hint: $$ \frac{\sin(x-y)}{x^3-y^3} = \frac{\sin(x-y)}{x-y} \cdot \frac{x-y}{x^3-y^3} = \frac{\sin(x-y)}{x-y} \cdot \frac{1}{x^2 + xy + y^2} $$ Use also the fact that $\frac{\sin(z)}{z} \to 1$ when $z \to 0$.

Answer 2

Let $L$ be the limit in question, we have: $x^2+xy+y^2 \ge 0, \forall x,y \in \mathbb{R}\implies L = +\infty$ because the factor $\dfrac{\sin(x-y)}{x-y}$ tends to $1$ when $(x,y) \to (0,0)$ .

Answer 3

Look at a sequence:

$$x_{n}=\frac{2}{n},\quad y_{n}=\frac{1}{n}$$,

Then $(x_{n},y_{n})\rightarrow (0,0)$ as $n\rightarrow\infty$

Shove this is to your limit and see what happens.