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

I was asked to find limit of $$f(x,y)= \frac{\sin^2 (x-y)}{|x|+|y|}$$ as $$(x,y)$$ approaches $$(0,0)$$. I have an idea that it tends to $$0$$, but can't find a suitable way to proceed.Please help with any suggestions to proceed.

as per suggestions I tried by converting the given function to polar co-ordinates. so $$\frac {(x-y)^2}{|x|+|y|}$$. which gives $$r\cdot \frac{(\cos(t)-\sin(t))^2}{(\cos(t)+\sin(t))}$$. On using the limit as r goes to zero, we get answer as zero. Is there anything else to be taken care in this case?

• What is $modx$ ? Please. Nov 7 '18 at 13:39
• @dmtri I suppose $|*|$. Why did you rollback the editing?
– user
Nov 7 '18 at 13:40
• Please show your work and effort here and try to use MATHJAX
– user
Nov 7 '18 at 13:42
• yes I meant modx for what you have written above. Nov 7 '18 at 13:44
• @gimusi, sorry it is the first time for me to see that absolute value is $modx$ ... Nov 7 '18 at 13:44

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

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.

• I did arrive at this . But is not sure how to proceed further. Nov 7 '18 at 13:35
• @samsoft What about polar coordinates for example?
– user
Nov 7 '18 at 13:35
• Then should the limit be like as r tend to zero Nov 7 '18 at 13:37
• @samsoft Yes of course, by polar coordinates we find an expression $f(x,y)=g(r,\theta)$ and if we show that $g(r,\theta) \to L$ as $r \to 0\quad \forall \theta$ then $f(x,y) \to L$.
– user
Nov 7 '18 at 13:39
• So if I use the same, I will be able to get ' r times some function of theta' which should then become zero, right? Nov 7 '18 at 13:41

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 $$(x,y) \to (0,0)$$.

Attempt:

$$|\sin^2(x-y)| \le |x-y|^2$$ , $$x,y$$ real.

$$|\dfrac{ \sin^2 (x-y)}{|x|+|y|}|\le \dfrac{|x-y|^2}{|x|+|y|} \le$$

$$\dfrac{ (|x|+|y|)^2}{|x|+|y|}= |x+y| \le$$

$$|x|+|y| = \sqrt{x^2} +\sqrt{y^2}\le$$

$$2\sqrt{x^2+y^2}.$$

Chose $$\delta =\epsilon/2$$.

• I don't get the inequality used in here: √x^2+√y2 ≤ 2√.(x2+y2) Nov 7 '18 at 16:13
• sam.$\sqrt{x^2} \le \sqrt{x^2+y^2}$, have made it bigger by adding $y^2$ (under the square root).Same with $\sqrt{y^2}\le \sqrt{y^2+x^2}$, adding positive $x^2$ under the square root.; used: √ is an increasing fct.means √4 <√5, for example. Is this ok? Nov 7 '18 at 16:30
• yeah i got it , thank you Nov 7 '18 at 17:26
• Sam.A pleasure. Nov 7 '18 at 17:45

Let $$x=r\cos\theta, y=r\sin\theta$$, we have \begin{align} \displaystyle\lim_{(x,y)\to(0,0)}\cfrac {\sin^2(x-y)}{|x|+|y|}&=\lim_{(x,y)\to(0,0)}\cfrac {(x-y)^2}{|x|+|y|}\\ &=\lim_{r\to 0}\cfrac{r(\cos\theta-\sin\theta)^2}{|\cos\theta|+|\sin\theta|}\\ \end{align}

We assert $$\cfrac{(\cos\theta-\sin\theta)^2}{|\cos\theta|+|\sin\theta|}$$ has upper bound $$1$$ and lower bound $$0$$, thus it's finite. So the RHS evaluate to $$0$$.