# 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.

• Is it really undefined? WRA says that the limit is equal to $1$. – Ethan Hunt Mar 26 '16 at 19:03
• I think you can only approach via paths that lie in the domain of the function. You just need to make sure there exists at least one such path (so it is not isolated point), and that all such paths approach to the $1$. – Sil Mar 26 '16 at 19:29
• The answer depends on your book's definition. Some people insist that the function be defined in a deleted neighbourhood of $(0,0)$. Others don't. – André Nicolas Mar 26 '16 at 19:40
• André is right. The only way this question can be given a definite answer is if you tell us the exact definition of limit that you are using. – Hans Lundmark Mar 26 '16 at 20:41
• The function must be defined in the a certain neighborhood of (0,0) with the exception of (0,0). Can you give me a little explanation on how does this information affects the result ? – UUKKS Mar 27 '16 at 16:05

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

$$\lim_{(x,y) \to (0,0)} \hat{f}(x,y) = 1.$$

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.

• So the limit is 1 even if the path x=y is undefined ? – UUKKS Mar 26 '16 at 19:12

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

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