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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?

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

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

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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 $\delta =\epsilon/2$.

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  • $\begingroup$ I don't get the inequality used in here: √x^2+√y2 ≤ 2√.(x2+y2) $\endgroup$
    – sam soft
    Nov 7 '18 at 16:13
  • $\begingroup$ 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? $\endgroup$ Nov 7 '18 at 16:30
  • $\begingroup$ yeah i got it , thank you $\endgroup$
    – sam soft
    Nov 7 '18 at 17:26
  • $\begingroup$ Sam.A pleasure. $\endgroup$ Nov 7 '18 at 17:45
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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 $0$.

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