# two variable limit

• May 21st 2010, 06:20 AM
venturozzaccio
two variable limit
Hi! I am new and I can't solve this limit:

limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)

Can anyone give me a hand?
• May 21st 2010, 06:35 AM
Prove It
Quote:

Originally Posted by venturozzaccio
Hi! I am new and I can't solve this limit:

limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^6|+|y^4|)

Can anyone give me a hand?

Try converting to polars:

$\displaystyle \lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$

$\displaystyle = \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$

$\displaystyle = \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$

$\displaystyle = \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$

$\displaystyle = \frac{0}{|\sin^4{\theta}|}$

$\displaystyle = 0$.
• May 21st 2010, 07:45 AM
venturozzaccio
sorry I make a mistake writing.

the real one is :

limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^2|+|y^4|)
• May 21st 2010, 12:17 PM
AllanCuz
Quote:

Originally Posted by Prove It
Try converting to polars:

$\displaystyle \lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$

$\displaystyle = \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$

$\displaystyle = \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$

$\displaystyle = \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$

$\displaystyle = \frac{0}{|\sin^4{\theta}|}$

$\displaystyle = 0$.

This is really nice (Clapping)
• May 21st 2010, 01:16 PM
venturozzaccio
yes but about the revised question?
• May 21st 2010, 08:25 PM
Prove It
Use the exact same process. You will still be able to cancel enough $\displaystyle r$'s to be able to evaluate the limit.
• May 22nd 2010, 02:54 AM
venturozzaccio
so the result is still 0 ?
• May 22nd 2010, 04:58 AM
Prove It
Quote:

Originally Posted by venturozzaccio
so the result is still 0 ?

Correct.
• May 29th 2010, 05:00 AM
venturozzaccio
ok. but if the limit is

limit x,y->(0,0) $\displaystyle xy^2 / (x^2+y^4)$

then I have $\displaystyle 0cos()sen^2() /( cos^2+0^2cos^4())$

And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?
• May 29th 2010, 05:48 AM
Prove It
Quote:

Originally Posted by venturozzaccio
ok. but if the limit is

limit x,y->(0,0) $\displaystyle xy^2 / (x^2+y^4)$

then I have $\displaystyle 0cos()sen^2() /( cos^2+0^2cos^4())$

And following the previous example this should be 0 again. But this time the limit doesn't exist. where I make a mistake?

I get that the limit is $\displaystyle 0$...

$\displaystyle \lim_{(x, y) \to (0, 0)}\frac{xy^2}{x^2 + y^4} = \lim_{r \to 0}\frac{r\cos{\theta}\,r^2\sin^2{\theta}}{r^2\cos^ 2{\theta} + r^4\sin^4{\theta}}$

$\displaystyle = \lim_{r \to 0}\frac{r^3\cos{\theta}\sin^2{\theta}}{r^2(\cos^2{ \theta} + r^2\sin^4{\theta})}$

$\displaystyle = \lim_{r \to 0}\frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta } + r^2\sin^4{\theta}}$

$\displaystyle = \frac{0}{\cos^2{\theta}}$

$\displaystyle = 0$.
• May 29th 2010, 06:10 AM
venturozzaccio
I think it's not true. Just because if you set x=y the limit is 0. But if you set x=y^2 then the limit is 1/2. So limit change according the way you approach to (0,0) and for this it doesn't exist.