# Does the following limit exist $\lim_{(x,y)\rightarrow (0,0)}\frac{\sin(xy)}{y}$

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 $$y=0.$$

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

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.

• The idea is just fine but you also need to address the (trivial) case when we are approaching $(0,0)$ by the line $x=0$. – PierreCarre May 16 at 17:32
• Thanks @PierreCarre – user376343 May 16 at 17:50

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 $$\lim_{(x,y)\to(0,0)}{\sin (xy)\over y}=\lim_{(x,y)\to(0,0)}\bigl(x\>{\rm sinc}(xy)\bigr)=0\cdot1=0\ .$$

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

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