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Can someone help me with this limit? $$ \lim_{(x,y)\to (0,0)} \frac{\sin(x-y)}{x+y} $$ Over the domain $x>0$, $y>0$.

I think the result is $0$, but I can't conclude any estimation due to $|x+y|$ in the denominator.

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    $\begingroup$ What do you mean by "help you with the limit?" Asking for help implies you are fully engaged and have worked on it and you include your effort, whether it's successful or not. It also means you specify what specifically you need help with. Else, you are using "Can someone help me with this..." to mean "Can someone do this for me", in which case anyone "doing it for you" is no longer helping you. Just please be clear: do you want help so you can solve this, or do you want someone to do it for you? If you want the latter, that's not this site's purpose. $\endgroup$ – amWhy Apr 1 at 21:53
  • $\begingroup$ @amWhy I'm sorry for that. I was pretty sure about my result and with this question I thought I would need only some suggestions for handling $|x+y|$. Next time I will be more clear about my doubts and I will show my attempts to reach the solution. $\endgroup$ – DOmonoXYLEDyL Apr 2 at 7:08
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HINT:

$$\frac{\sin(x-y)}{x+y}=\left(\frac{\sin(x-y)}{x-y}\right)\left(\frac{x-y}{x+y}\right)\tag1$$

The first parenthetical term on the right-hand side of $(1)$ approaches $1$. Does the second parenthetical term have a limit as $(x,y)\to (0^+,0^+)$?

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It does not exist. If we take the limit about the line $y=mx$, the result depends on $m$, which should not happen if the limit did exist.

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  • $\begingroup$ Limits aren't discontinuous. Functions may be discontinuous. Here, we say the limit does not exist. And it does not exist because from various paths of approaching $(0, 0)$ via $y= mx$ gives different results depending on $m$. $\endgroup$ – amWhy Apr 1 at 22:11
  • $\begingroup$ Thanks! That was sloppy, I admit. What I meant was that the function $\frac{\sin(x-y)}{x+y}$ was discontinuous. $\endgroup$ – Ishan Deo Apr 1 at 22:17
  • $\begingroup$ No problem! I figured that that's what you meant, but the "it is discontinuous" seemed to address the limit, per the question post. $\endgroup$ – amWhy Apr 1 at 22:19
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Consider $y=x$ and $y=x^2$. As $(x,y)\to(0,0)$ along the first curve, we get $0$. Along the second, we get $1$, by Lhopital.

Therefore the limit doesn't exist.

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Along the path $x=0, y\rightarrow 0$ limit value of the given expression is $-1$ and along the path $y=0, x\rightarrow 0$ limit value is 1. Different paths yield different limit. So limit does not exist.

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