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:
$\displaystyle \lim_{(x, y) \to (0,0)}\left(\frac{x^2y^4}{|x^6| + |y^4|}\right) = \lim_{r \to 0}\left[\frac{(r\cos{\theta})^2(r\sin{\theta})^4}{|(r\cos{ \theta})^6| + |(r\sin{\theta})^4|}\right]$
$\displaystyle = \lim_{r \to 0}\left[\frac{r^6\cos^2{\theta}\sin^4{\theta}}{r^6|\cos^6{ \theta}| + r^4|\sin^4{\theta}|}\right]$
$\displaystyle = \lim_{r \to 0}\left[\frac{r^2\cos^2{\theta}\sin^4{\theta}}{r^2|\cos^6{ \theta}| + |\sin^4{\theta}|}\right]$
$\displaystyle = \frac{0^2\cos^2{\theta}\sin^4{\theta}}{0^2|\cos^6{ \theta}| + |\sin^4{\theta}|}$
$\displaystyle = \frac{0}{|\sin^4{\theta}|}$
$\displaystyle = 0$.
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 $\displaystyle 0$...
$\displaystyle \lim_{(x, y) \to (0, 0)}\frac{xy^2}{x^2 + y^4} = \lim_{r \to 0}\frac{r\cos{\theta}\,r^2\sin^2{\theta}}{r^2\cos^ 2{\theta} + r^4\sin^4{\theta}}$
$\displaystyle = \lim_{r \to 0}\frac{r^3\cos{\theta}\sin^2{\theta}}{r^2(\cos^2{ \theta} + r^2\sin^4{\theta})}$
$\displaystyle = \lim_{r \to 0}\frac{r\cos{\theta}\sin^2{\theta}}{\cos^2{\theta } + r^2\sin^4{\theta}}$
$\displaystyle = \frac{0}{\cos^2{\theta}}$
$\displaystyle = 0$.