How to find $ \lim_{(x,y)\to(0,0)} \frac{\sin^2 (x-y)}{|x|+|y|} $?

Solution 1:

HINT

We have that

$$\frac{\sin^2 (x-y)}{|x|+|y|}=\frac{\sin^2 (x-y)}{(x-y)^2}\cdot \frac{(x-y)^2}{|x|+|y|}$$

and by $t=x-y \to 0$

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

then we need to evaluate $\frac{(x-y)^2}{|x|+|y|}\to ?$.

Added after editing

For the latter, by polar coordinates we obtain

$$r\cdot \frac{(\cos t-\sin t)^2}{ |\cos t|+ |\sin t|}$$

and the key point here is that the denominator $|\cos t|+ |\sin t|$ is bounded and never equal to zero therefore for some $c>0$

$$0\le r\cdot \frac{(\cos t-\sin t)^2}{ |\cos t|+ |\sin t|}\le r\cdot c$$

and we can conclude by squeeze theorem.

Solution 2:

With $|\sin t | \le |t|$ we get $\frac{|\sin (x-y)|}{|x|+|y|} \le 1$, hence

$0 \le \frac{\sin^2 (x-y)}{|x|+|y|} \le |\sin (x-y)|$.

This gives $\frac{\sin^2 (x-y)}{|x|+|y|} \to 0$ as $(x,y) \to (0,0)$.