1. ## Limits in R³

For limits such as

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin(x^2+y^2)}$

Can we convert to polar co-ordinates?

So since $\displaystyle r^2=x^2+y^2$ this implies that as $\displaystyle (x,y)\to(0,0)\Rightarrow{r\to{0}}$

So we would have

$\displaystyle \lim_{r\to{0}}\frac{r^2}{\sin\left(r^2\right)}=\li m_{r\to{0}}\frac{r^2}{r^2-\frac{r^6}{3!}+...}=1$

I wonder if this works, for my book doesn't say anything about it, but theoretcally it makes sense to me.

Anyone know?

Mathstud.

2. Hello,

Originally Posted by Mathstud28
For limits such as

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin(x^2+y^2)}$

Can we convert to polar co-ordinates?

So since $\displaystyle r^2=x^2+y^2$ this implies that as $\displaystyle (x,y)\to(0,0)\Rightarrow{r\to{0}}$

So we would have

$\displaystyle \lim_{r\to{0}}\frac{r^2}{\sin\left(r^2\right)}=\li m_{r\to{0}}\frac{r^2}{r^2-\frac{r^6}{3!}+...}=1$

I wonder if this works, for my book doesn't say anything about it, but theoretcally it makes sense to me.

Anyone know?

Mathstud.
Yes you can. Converting into polar coordinates lets you deal with all the possible directions of x & y to 0 & 0.

You can also note that $\displaystyle \lim_{r \to 0} \frac{r^2}{\sin(r^2)}=\frac 11=1$

3. Originally Posted by Moo
Hello,

Yes you can. Converting into polar coordinates lets you deal with all the possible directions of x & y to 0 & 0.
Thank you Moo!

So the the obvious extension would be

$\displaystyle \lim_{(x,y,z)\to(0,0,0)}\frac{\sqrt{x^2+y^2+z^2}}{ \sin\left(\sqrt{x^2+y^2+z^2}\right)}$

Corresponds to

$\displaystyle \lim_{r\to{0}}\frac{r}{\sin(r)}=1$ by spherical coordinates?

Does this work when converting to any co-ordinate system?

4. Originally Posted by Mathstud28
For limits such as

$\displaystyle \lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin(x^2+y^2)}$

Can we convert to polar co-ordinates?

So since $\displaystyle r^2=x^2+y^2$ this implies that as $\displaystyle (x,y)\to(0,0)\Rightarrow{r\to{0}}$

So we would have

$\displaystyle \lim_{r\to{0}}\frac{r^2}{\sin\left(r^2\right)}=\li m_{r\to{0}}\frac{r^2}{r^2-\frac{r^6}{3!}+...}=1$

I wonder if this works, for my book doesn't say anything about it, but theoretcally it makes sense to me.

Anyone know?

Mathstud.
you don't need polar coordinates ... if $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ is such that $\displaystyle \lim_{\bold{x}\to\bold{0}}f(\bold{x})=0,$ then $\displaystyle L=\lim_{\bold{x}\to\bold{0}} \frac{\sin(f(\bold{x}))}{f(\bold{x})}=1,$ where

$\displaystyle \bold{x}=(x_1, \ ... \ ,x_n), \ \bold{0}=(0, \ ... \ , 0).$ because you can just put $\displaystyle f(\bold{x})=u.$ then since $\displaystyle u \rightarrow 0$ as $\displaystyle \bold{x} \rightarrow \bold{0},$ we'll have that

$\displaystyle L=\lim_{u\to0} \frac{\sin u}{u} = 1.$

5. Originally Posted by NonCommAlg
you don't need polar coordinates ... if $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ is such that $\displaystyle \lim_{\bold{x}\to\bold{0}}f(\bold{x})=0,$ then $\displaystyle L=\lim_{\bold{x}\to\bold{0}} \frac{\sin(f(\bold{x}))}{f(\bold{x})}=1,$ where

$\displaystyle \bold{x}=(x_1, \ ... \ ,x_n), \ \bold{0}=(0, \ ... \ , 0).$ because you can just put $\displaystyle f(\bold{x})=u.$ then since $\displaystyle u \rightarrow 0$ as $\displaystyle \bold{x} \rightarrow \bold{0},$ we'll have that

$\displaystyle L=\lim_{u\to0} \frac{\sin u}{u} = 1.$
But isn't that exactly what I did, I let $\displaystyle r=x^2+y^2$ regardless if it was polar coordinates, and then as $\displaystyle (x,y)\to(0,0)$ we have that $\displaystyle r\to{0}$, so whether or not I called it polar coordiates wasn't it basically teh same concept?

6. Originally Posted by Mathstud28
But isn't that exactly what I did, I let $\displaystyle r=x^2+y^2$ regardless if it was polar coordinates, and then as $\displaystyle (x,y)\to(0,0)$ we have that $\displaystyle r\to{0}$, so whether or not I called it polar coordiates wasn't it basically teh same concept?
you actually did use polar coordinates! haha ... anyway, my point was that it's true in general for any function

$\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ which satisfies the condition $\displaystyle \lim_{\bold{x}\to\bold{0}}f(\bold{x})=0.$

7. Originally Posted by NonCommAlg
you actually did use polar coordinates! haha ... anyway, my point was that it's true in general for any function

$\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ which satisfies the condition $\displaystyle \lim_{\bold{x}\to\bold{0}}f(\bold{x})=0.$

Let $\displaystyle f(x,y,z,\cdots)\to(a,b,c,\cdots)=0$

Then

$\displaystyle \lim_{(x,y,z,\cdots)\to(a,b,c,\cdots)}\frac{\sin(f (x,y,z,\cdots))}{f(x,y,z,\cdots)}$

We can say

Let $\displaystyle f(x,y,z,\cdots)=u$

So as $\displaystyle f(x,y,z,\cdots)\to(a,b,c,\cdots)=0\Rightarrow{u\to (a,b,c,\cdots)=0}$

Giving us

$\displaystyle \lim_{u\to{0}}\frac{\sin(u)}{u}$?

8. let's do a very general case:

Claim: suppose $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ such that $\displaystyle \lim_{\bold{x}\to\bold{a}}f(\bold{x})=b,$ and $\displaystyle \lim_{x\to b}g(x)=L.$ then $\displaystyle \lim_{\bold{x}\to\bold{a}}g(f(\bold{x}))=L.$

Proof: let $\displaystyle \epsilon > 0$ be given. since $\displaystyle \lim_{x\to b}g(x)=L,$ there exists $\displaystyle \delta_0 > 0$ such that whenever $\displaystyle |x-b| < \delta_0,$ we'll have $\displaystyle |g(x)-L| < \epsilon.$

on the other hand, since $\displaystyle \lim_{\bold{x}\to\bold{a}}f(\bold{x})=b,$ there exists $\displaystyle \delta > 0$ such that whenever $\displaystyle |\bold{x}-\bold{a}| < \delta,$ we will have $\displaystyle |f(\bold{x})-b| < \delta_0.$

now it's clear that if $\displaystyle |\bold{x}-\bold{a}| < \delta,$ then $\displaystyle |g(f(\bold{x}))-L| < \epsilon. \ \ \ \square$