1. [SOLVED] Multivariable limit (x,y)&gt;(0,0)

Find limit of $\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^2+y^2}$

Does it go like this?
$\lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho \sin \theta}{\rho^{2}}=\lim_{\rho \rightarrow 0}\rho \cdot \cos^{2}\theta \sin \theta=?$

2. Originally Posted by Revy
Find limit of $\lim_{(x,y)\rightarrow(0,0)}\frac{x^2y}{x^2+y^2}$

Does it go like this?
$\lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho \sin \theta}{\rho^{2}}=\lim_{\rho \rightarrow 0}\rho \cdot \cos^{2}\theta \sin \theta=?$
Looks good to me.

So what happens as $\rho \to 0$?

3. Limit is 0
I'm not sure after all I'm extremely bad at math

4. Yes.

Though I found a mistake in your simplification.

It should be

$\lim_{\rho \to 0}\rho ^2\cos^2{\theta}\sin{\theta}$.

And I'd say you can't be THAT bad at maths if you are doing multivariable calculus...

5. you did $\lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho^{2} \sin \theta}{\rho^{2}}$?
why is it so? After all $x=\rho \cdot \cos\theta$ and $y=\rho \cdot \sin\theta$

After trying to understand various problems for 10hours, almost everyone would start to understand something

In other thread I found that $\lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)$ => $y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.$
In what way does 2 appear?

6. Originally Posted by Revy
you did $\lim_{\rho \rightarrow 0}\frac{\rho^{2}\cos^{2}\theta\cdot\rho^{2} \sin \theta}{\rho^{2}}$?
why is it so? After all $x=\rho \cdot \cos\theta$ and $y=\rho \cdot \sin\theta$

After trying to understand various problems for 10hours, almost everyone would start to understand something

In other thread I found that $\lim_{(x,y)\rightarrow(0,0)}y^2\ln(x^2+y^2)$ => $y^2 \ln(x^2+y^2)=2r^2 \sin^2 \theta \ln r.$
In what way does 2 appear?
Haha you are right, for some reason I misread and thought you only need to divide by $\rho$, not $\rho^2$.

Never mind, you got the answer right.

$\lim_{(x, y) \to (0, 0)}y^2\ln{(x^2 + y^2)} = \lim_{r \to 0}[(r\sin{\theta})^2 \ln{(r^2)}]$

$= \lim_{r \to 0}[r^2\sin^2{\theta} \ln{(r^2)}]$.

Now remember the logarithm rule $\log{(n^p)} = p\log{n}$?

That means you can take the power out as a factor...

So it becomes

$\lim_{r \to 0}[r^2 \sin^2{\theta} \cdot 2\ln{r}]$

$= \lim_{r \to 0} [2r^2 \sin^2{\theta} \ln{r}]$.

Notice though that you can't solve this using direct substitution, since $\log{0}$ is undefined. It will tend to the indeterminate $0\cdot (-\infty)$. This one requires a bit more thought.

7. The crucial point, for both of these, is that once you have put the problem in polar coordinates, the limit, as r goes to 0, does not depend on $\theta$! Since r itself determines "closeness" to (0,0), that is what you must have in order that you do not get different limits approaching (0,0) along different paths- in order that the limit exist.