B Is this the right set mapping notation for a limit in two variables?

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The discussion centers on the validity of a set mapping notation for a limit involving two variables, specifically the limit of the function y*sin(x/y) as (x,y) approaches (π/3, 2). The proposed notation indicates that the limit evaluates to 1, and the user seeks confirmation of its unambiguity and correctness. Participants emphasize the importance of continuity and the epsilon-delta definition in proving the limit, while also noting that the function is a composition of continuous functions. There is a consensus that the approach to proving the limit could be refined, particularly regarding the handling of inequalities and the definition of neighborhoods. Overall, the discussion highlights the complexities of multivariable limits and the necessity for precise notation and proof techniques.
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I want to know if my attempt is well-defined, unambiguous.
$$\displaystyle\lim_{(x,\,y)\rightarrow{(\pi/3,\,2)}}{y\sin\bigg(\cfrac{x}{y}\bigg)}=2\sin\bigg(\cfrac{\pi}{6}\bigg)=1$$

Set mapping notation I've attempted:

$$\mathbb{R}^2\,\,\,\longrightarrow{\quad{\mathbb{R}}}$$
$$f(x,y)\,\longrightarrow{1}$$

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In this case, it is unambiguous since all possible paths in a local neighborhood lead to the same point. This in general not automatically the case.
 
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mcastillo356 said:
TL;DR Summary: I want to know if my attempt is well-defined, unambiguous.

lim(x,y)→(π/3,2)ysin⁡(xy)=2sin⁡(π6)=1

Set mapping notation I've attempted:

R2⟶R
f(x,y)⟶1

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For me, the notation f(x,y)⟶1 means that you have defined a new constant function z⟶1 where z=f(x,y). You have defined the function composition (x,y)⟶f(x,y)⟶1.
 
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I'm working on an ##\epsilon/\delta## proof of the limit of ##f(x,\,y)=y\sin\bigg (\cfrac{x}{y}\bigg)##. It will take me a while just to arrange an attempt.

Gavran said:
For me, the notation f(x,y)⟶1 means that you have defined a new constant function z⟶1 where z=f(x,y). You have defined the function composition (x,y)⟶f(x,y)⟶1.

Yes.

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mcastillo356 said:
I'm working on an ##\epsilon/\delta## proof of the limit of ##f(x,\,y)=y\sin\bigg (\cfrac{x}{y}\bigg)##. It will take me a while just to arrange an attempt.



Yes.

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##\left(\mathbb{R}\setminus \{0\},\cdot\right)## is a topological group, so inversion and multiplication are continuous, as is the sine function. Hence, your function is a sequence of continuous functions. I would use the topological definitions with open sets and write it down in those terms before I translated it into the ##\varepsilon-\delta## world. Especially the product topology, which you need for the multiplication in between, can be tricky in the latter.
 
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mcastillo356 said:
I'm working on an ##\epsilon/\delta## proof of the limit of ##f(x,\,y)=y\sin\bigg (\cfrac{x}{y}\bigg)##. It will take me a while just to arrange an attempt.



Yes.

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I'm not sure what an epsilon-delta proof of that would achieve. You can use epsilon-delta to prove basic theorems, then use those theorems to confirm continuity for more complex functions.
 
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$$\displaystyle\int_{(x,\,y)\rightarrow{(\pi/3,\,2)}}{y\sin\bigg(\cfrac{x}{y}\bigg)}=1$$
means that
$$\forall{\epsilon>0},\,\exists{\delta>0}$$
such that whenever
$$0<\sqrt{(x-\pi/3)^2+(y-2)}<\delta$$
then
$$\bigg | y\sin\bigg(\cfrac{x}{y}\bigg)-1\bigg |<\epsilon$$
Proof
Given ##\epsilon>0##, we must choose ##\delta{(\epsilon)}## to be something.

Suppose
$$0<\sqrt{(x-\pi/3)^2+(y-2)}<\delta$$

Check

$$\bigg | y\sin\bigg(\cfrac{x}{y}\bigg)-1\bigg |<|y|\Big | y\sin\bigg(\cfrac{x}{y}\bigg)-1\Big |<|y|\Big | y\sin\bigg(\cfrac{x}{y}\bigg)\Big |<|y|\bigg |\cfrac{x}{y}\bigg |=|x|\leq{\sqrt{(x-\pi/3)^2+(y-2)^2}}$$

$$y\sin\bigg (\cfrac{x}{y}\bigg )=t\Rightarrow{\cfrac{x}{y}} =\arcsin{(t)}+2\pi\cdot{K},\quad{K\in{\mathbb{K}}}$$

set ##k=0##

Then ##\epsilon=\delta## is the choice

fresh_42 said:
In this case, it is unambiguous since all possible paths in a local neighborhood lead to the same point. This in general not automatically the case.

Did I by the way define a neighborhood?

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Except that the integral is probably a limit, the idea is ok.

We need to show that the preimage of an open set is an open set.

Set ##f(x,y)=y\sin\left(\dfrac{x}{y}\right)## and ##U(p,r)=\{q\in \mathbb{R}^n\,|\,\|q-p\|<r\}.## An open set in the domain around ##1## contains an open ball ##U(1,\varepsilon)=\{p\in \mathbb{R}\,|\,\|p-1\|<\varepsilon\},## and we want to show that ##f^{-1}\left(U(1,\varepsilon)\right)=\{(x,y)\in \mathbb{R}^2\,|\,f(x,y)\in U(1,\varepsilon)\}## is open. We therefore show that it contains an open ball ##U((\pi/3,2),\delta).## This means that we have to show that ##f\left(U((\pi/3,2),\delta)\right)\subseteq U(1,\varepsilon).##

Now we choose a point ##(x,y)\in U((\pi/3,2),\delta)## which means that $$\|(x,y)-(\pi/3,2)\|=\sqrt{(x-\pi/3)^2+(y-2)^2} <\delta\quad(1)$$
is given, and we have to show that
$$
\|f(x,y)-1\|=\left|y\sin\left(\dfrac{x}{y}\right)-1\right|<\varepsilon\quad (2).
$$

So we have to show the last inequality by using the first. However, I do not follow your inequalities. Yes, we may assume that ##|y|>1,## but this has to be noted for our choice of ##\delta## or must be proven. And then, the first inequality is ##\le ## and not strict, since the sine term could be zero. Regardless of those tiny mistakes, I do not see where the ##-1## and the sine in the next steps go.

The problem is to show that ##x/y## is near ##\pi/6## only using the Euclidean distance. I would show that ##f## is continuous, partially differentiable, and continuity follows from that. I have tried to use the series expansion, but with no luck. One has to get rid of ##x/y## first.
 
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PeroK said:
I'm not sure what an epsilon-delta proof of that would achieve. You can use epsilon-delta to prove basic theorems, then use those theorems to confirm continuity for more complex functions.
In fact, the cartesian coordinates are continous, and sine function too; this means the value of the limit at any points ##(x,\,y)## will be the same as the function's value. In this case:

$$\displaystyle\lim_{(x,\,y)\rightarrow{(a,\,b)}}{f(x,\,y)g(x,\,y)}=LM$$

##L## and ##M## being the values of ##y## and ##\sin{\Big (\cfrac{x}{y}\Big )}##

My attempt to prove the limit in a ##\epsilon/\delta## fashion is not wise at all.
 
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