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

[mcastillo356]

TL;DR

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:

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

No preview.

 

[fresh_42]

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.

 

[Gavran]

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

No preview.

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.

 

[mcastillo356]

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.

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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[fresh_42]

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.

No preview

$\left(\mathbb{R}\setminus \{0\},\cdot\right)$ 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.

 

[PeroK]

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.

No preview

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.

 

[mcastillo356]

$$\displaystyle\int_{(x,\,y)\rightarrow{(\pi/3,\,2)}}{y\sin\bigg(\cfrac{x}{y}\bigg)}=1$$
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

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?

No preview

 

[fresh_42]

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.

 

[mcastillo356]

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.