Limit $\lim\limits_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}$ I'm trying to understand the following limit:
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}$$
The $\lim_{(x,y)\to (0,0)} f(x,y)$ is undefined. Why it is not equal to $1$?
Let's suppose $t = x^2-y^2$. Then as $(x,y)$ approaches $(0,0)$, $t$ approaches $0$, so 
$$\lim_{t\to 0} \frac{\sin t}{t} = 1 $$
I can see that the path $x=y$ is undefined,does this mean that the limit does not exist ?
I though that I needed to find two different defined paths that gives two different results to disprove that the limit exists.
If so this means that I can't really rely on the "t substitution" to determined the limit.
I tried to pass it into polar coordinates but It didn't help..
How can I determine a limit of a two variable function ? I can't check all the possibles paths like in one variable (left and right ).
I understood that the safest way is to use the squeeze theorem  or to pass the coordinates into polar coordinates and then use the squeeze theorem.
I would appreciate any sort of help in the matter ,
Thanks.
 A: You can use polar coordinates here. Set $x=r\cos\theta$, $y=r\sin\theta$, then notice that $x^2-y^2=r^2\cos 2\theta$. Then the limit becomes
$$\lim_{r \to 0} \frac{\sin (r^2\cos 2\theta)}{r^2\cos 2 \theta}.$$
Clearly you have to exclude the case $\theta=\pm \pi/4$ because $f$ is not defined there, even if you can try to extend it by continuity.
A: We have the inequality $z - z^3/6 \leqslant \sin z \leqslant z$ for $z > 0$.
Hence, for $x  > y$
$$1 - \frac{(x^2 - y^2)^2}{6} \leqslant \frac{\sin (x^2 - y^2)}{x^2 - y^2} \leqslant 1,$$
and by the squeeze theorem the limit is $1$ as $(x,y) \to (0,0)$ with $x > y.$
Similarly, for $y  > x$
$$1 - \frac{(y^2 - x^2)^2}{6} \leqslant \frac{\sin (y^2 - x^2)}{y^2 - x^2} = \frac{\sin (x^2 - y^2)}{x^2 - y^2} \leqslant 1,$$
and by the squeeze theorem the limit is $1$ as $(x,y) \to (0,0)$ with $y > x.$
The inequality also shows that as $x \to y$ with $y$ fixed,
$$\lim_{x \to y} \frac{\sin (x^2 - y^2)}{x^2 - y^2} = 1,$$
and, although the function as written above is undefined for $x = y,$ it can be extended continuously to $1$ on that line.
Technically, with $f(x,y) = \sin(x^2 - y^2) /( x^2 - y^2),$ you would say
$$\lim_{(x,y) \to (0,0), x \neq y} f(x,y) = 1,$$
and $f$ can be continuously extended to a function $\hat{f}$ on $\mathbb{R}^2$ such that
$$\lim_{(x,y) \to (0,0)} \hat{f}(x,y) = 1.$$
A: Recall from elementary geometry that the sine function satisfies the inequalities 
$$|\theta\cos(\theta)|\le |\sin(\theta)|\le |\theta|$$
for $|\theta|\le \pi/2$. 
Letting $\theta =x^2-y^2$ we can write
$$|\cos(x^2-y^2)|\le\left|\frac{\sin(x^2-y^2)}{x^2-y^2}\right|\le|1|$$
whereupon applying the squeeze theorem and exploiting the evenness of $\frac{\sin(z)}{z}$ yields the limit
$$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2-y^2)}{x^2-y^2}=1$$
