1. ## Finding the limit

Using polar corrdinates, evaluate $\lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^2+y^3}$

so $x=r\sin\theta$, $y=r\cos\theta$

$\frac{x^3+y^3}{x^2+y^3}=\frac{r(\sin^3\theta +\cos^3\theta)}{\sin^2\theta +\cos^2\theta}$

Since $\left|\frac{r(\sin^3\theta +\cos^3\theta)}{\sin^2\theta +\cos^2\theta}\right|\leq r$ hence limit=0

Can I ask why is that so?

when x,y goes to 0 respectively, which variable goes to 0? r or $\theta$?

2. Originally Posted by noob mathematician
Using polar corrdinates, evaluate $lim_{(x,y)\rightarrow(0,0)} \frac{x^3+y^3}{x^2+\color{red}y^3}$
Did you notice that it is $y^3$ in the denominator?
Is that a typo?

If not, the limit does not exist.
Look at the path along the x-axis and the along the y-axis.

3. Do you mean

$\lim_{\small (x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y^2}?$

If so, then using the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$, we obtain

$\frac{x^3+y^3}{x^2+y^2}=\frac{r^3(\cos^3\theta+\si n^3\theta)}{r^2(\cos^2\theta+\sin^2\theta)}=\frac{ r^3(\cos^3\theta+\sin^3\theta)}{r^2\cdot 1}=r(\cos^3\theta+\sin^3\theta).$

Now notice that $r$ is the variable that goes to $0$ (as we're approaching the origin), and

$|\cos^3\theta+\sin^3\theta|\le|\cos^3\theta|+|\sin ^3\theta|\le\cos^2\theta+\sin^2\theta=1.$

4. Originally Posted by Plato
Did you notice that it is $y^3$ in the denominator?
Is that a typo?

If not, the limit does not exist.
Look at the path along the x-axis and the along the y-axis.
Oh yeah it shd be y^2

Originally Posted by Scott H
Do you mean

$\lim_{\small (x,y)\rightarrow (0,0)}\frac{x^3+y^3}{x^2+y^2}?$

If so, then using the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$, we obtain

$\frac{x^3+y^3}{x^2+y^2}=\frac{r^3(\cos^3\theta+\si n^3\theta)}{r^2(\cos^2\theta+\sin^2\theta)}=\frac{ r^3(\cos^3\theta+\sin^3\theta)}{r^2\cdot 1}=r(\cos^3\theta+\sin^3\theta).$

Now notice that $r$ is the variable that goes to $0$ (as we're approaching the origin), and

$|\cos^3\theta+\sin^3\theta|\le|\cos^3\theta|+|\sin ^3\theta|\le\cos^2\theta+\sin^2\theta=1.$

Oh so it's actually the r goes to 0 and not the $\theta$?

Then by squeeze theorm we conclude it goes to 0 right?

5. Originally Posted by noob mathematician
Oh so it's actually the r goes to 0 and not the $\theta$?

Then by squeeze theorm we conclude it goes to 0 right?
Correct.

If $\theta$ goes to $0$, our point approaches the $x$-axis, but not necessarily the origin.