Question

Find
limit:

Answer 1

of?

Answer 2

\[\lim_{(x,y) \rightarrow (0,0)} \frac{ x^2y^3}{ \sqrt(x^2 +y^3) } \]

Answer 3

evaluate the limit by converting to polar coordinates

Answer 4

to convert into polar coordinates, use:
\[x=r*Cos(\theta)\]\[y=r*Sin(\theta)\]\[x^2+y^2=(r*Cos(\theta))^2+(r*Sin(\theta))^2=r^2Cos^2(\theta)+r^2Sin^2(\theta)\]\[x^2+y^2=r^2(Cos^2(\theta)+Sin^2(\theta))=r^2*1=r^2\]

Answer 5

I know how to convert it, but I'm a little stuck with the theta involved in the limit.

Answer 6

So that whole limit becomes:\[(r=\sqrt{x^2+y^2}=\sqrt{0^2+0^2}=0)\]
\[\lim_{r \rightarrow 0} \frac{ r^2Cos^2(\theta)r^3Sin^3(\theta)}{ \sqrt{r^2Cos^2(\theta) +r^3Sin^3(\theta)} }\]

Answer 7

Okay, I'm stuck. Was reading this, if it's of any help:
http://www.maths.adelaide.edu.au/michael.albanese/other/polar.pdf

@amistre64 mind helping us out here?

Answer 8

I got stuck when i reach this:
\[\lim_{r \rightarrow 0+} \frac{ r^5 \sin^22\theta \sin \theta }{ \sqrt(r(r+\sin \theta)) }\]
I tried expanding the sin sq and sin using product to sum, but it didn't help much.

Answer 9

\[\lim_{r \to 0} \frac{ r^2\cos^2(t)r^3\sin^3(t)}{ \sqrt{r^2\cos^2(t) +r^3\sin^3(t)} }\]
\[\lim_{r \to 0} \frac{ r^2\cos^2(t)r^3\sin^3(t)}{r \sqrt{\cos^2(t) +r\sin^3(t)} }\]
\[\lim_{r \to 0} \frac{ r^4\cos^2(t)\sin^3(t)}{\sqrt{\cos^2(t) +r\sin^3(t)} }\]
seems to me that just zeros out the top without hurting the bottom

Answer 10

ok, thanks!

Answer 11

i cant say how correct it is, but that does seem to play out like that

Answer 12

http://www.wolframalpha.com/input/?i=limit+%28x+to+0%2Cy+to+0%29+%2C+x%5E2y%5E3%2Fsqrt%28x%5E2%2By%5E3%29

it appears to be correct :)

Answer 13

I have the answer in the book, it's 0, i just needed to know why and how it is zero.

Answer 14

the offending part was the r attached to the underside .... by factoring out an r^2 we were able to remove its offensive behaviour

Answer 15

@bahrom7893 thanks for helping!

Answer 16

make sense, thanks!