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

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TL;DR Summary
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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