two variable limit

May 2010
7
1
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?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 2010
7
1
sorry I make a mistake writing.

the real one is :


limit x,y->(0,0) (( x^2 ) * ( y^4)) / (|x^2|+|y^4|)
 
Apr 2010
384
153
Canada
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)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Use the exact same process. You will still be able to cancel enough \(\displaystyle r\)'s to be able to evaluate the limit.
 
May 2010
7
1
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?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle 0\)...


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

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

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

\(\displaystyle = \frac{0}{\cos^2{\theta}}\)

\(\displaystyle = 0\).