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Does the following limit exist?

$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}$

The answer here is not correct in my opinion since it does not take under considerarion all the surface parts in which $y=0.$

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    $\begingroup$ $$y=0$$ makes no sence then in this case we get $$\frac{\sin(0)}{0}$$ $\endgroup$ – Dr. Sonnhard Graubner May 16 at 17:24
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    $\begingroup$ @DiegoMath How can $(1,y)$ be included when $(x,y)\to(0,0)$? $\endgroup$ – Shubham Johri May 16 at 17:32
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    $\begingroup$ @DiegoMath The limit does exist... your second path is not passing through $(0,0)$. $\endgroup$ – PierreCarre May 16 at 17:34
  • $\begingroup$ The expression under limit does not exceed $|x|$ in a neighborhood of $(0,0)$ hence by definition of limit the desired limit is $0$. $\endgroup$ – Paramanand Singh May 17 at 2:17
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Set $g(x)=\begin{cases} \frac{x\cdot \sin(xy)}{xy} \quad \text{if}\quad x\neq 0\\ 0 \quad \quad \quad \;\; \text{if}\quad x=0 \end{cases}$

Then $$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}=\lim_{(x,y)\rightarrow (0,0)}g(x)=0,$$ as we can make $\frac{\sin(xy)}{xy}$ as close to $1$ as we wish.

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    $\begingroup$ The idea is just fine but you also need to address the (trivial) case when we are approaching $(0,0)$ by the line $x=0$. $\endgroup$ – PierreCarre May 16 at 17:32
  • $\begingroup$ Thanks @PierreCarre $\endgroup$ – user376343 May 16 at 17:50
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There is a wonderful (entire) function $${\rm sinc}(t):=\left\{\eqalign{{\sin t\over t}\quad&(t\ne0)\cr 1\quad\ \ &(t=0)\cr}\right.$$ with ${\rm sinc}(0)=1$, having Taylor series $${\rm sinc}(t)=1-{1\over3!}t^2+{1\over5!}t^4-\ldots\ ,$$ and satisfying the identity $$t\>{\rm sinc}(t)=\sin t\qquad(t\in{\mathbb R})\ .$$ In terms of this function we have $${\sin (xy)\over y}={xy\>{\rm sinc}(xy)\over y}=x\>{\rm sinc}(xy)$$ for all points $(x,y)$ where the LHS is defined, i.e., for all $(x,y)$ with $y\ne0$. For this domain we can therefore say that $$\lim_{(x,y)\to(0,0)}{\sin (xy)\over y}=\lim_{(x,y)\to(0,0)}\bigl(x\>{\rm sinc}(xy)\bigr)=0\cdot1=0\ .$$

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$$\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y} = \lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{xy}x =(\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{xy})( \lim _{x\to 0} x)=1\times 0 =0$$

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Thanks for all the answers!

I came up with another proof:

Proof: Since there exists a $\delta$ neighborhood of $(0,0)$, $\mathcal{B_{\delta}(0,0)}$ , in which $|sin(xy)|<|xy|$ for every $(x,y) \in \mathcal{B_{\delta}(0,0)}$, we can write for every such $(x,y)$:

$|\frac{sin(xy)}{y}|=\frac{|sin(xy)|}{|y|}<\frac{|xy|}{|y|}=|x| \rightarrow 0$

What you think, is it ok?

Thanks!

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