# two variable limit

#### 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?

#### Prove It

MHF Helper
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:

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

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

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

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

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

$$\displaystyle = 0$$.

#### venturozzaccio

sorry I make a mistake writing.

the real one is :

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

#### AllanCuz

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)

#### venturozzaccio

yes but about the revised question?

#### Prove It

MHF Helper
Use the exact same process. You will still be able to cancel enough $$\displaystyle r$$'s to be able to evaluate the limit.

#### venturozzaccio

so the result is still 0 ?

MHF Helper

#### 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?

#### Prove It

MHF Helper
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 $$\displaystyle 0$$...

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

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

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

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

$$\displaystyle = 0$$.

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