Question

multivariable limit

Answer 1

\[\lim_{x,y \rightarrow (0,0)} \frac{ \sin(x^2+y^2) }{ 5x^2+5y^2 }\]

Answer 2

tried converting to polar and get

\[\frac{ \sin(r^2) }{ 5r^2 }\]

Answer 3

answer is 1/5

Answer 4

forgot how i did it originally.....

Answer 5

looks good! just notice that as \((x,y)\to(0,0)\), we have \(r\to 0\)

Answer 6

\[\lim_{x,y \rightarrow (0,0)} \frac{ \sin(x^2+y^2) }{ 5x^2+5y^2 } = \lim\limits_{r\to 0} \dfrac{\sin r^2}{5r^2}=?\]

Answer 7

0/0

Answer 8

recall the famous limit sinx/x

Answer 9

ah i see, lhoptials should work actually

Answer 10

wed have 2rcos(0)/10r = 2/10 = 1/5

Answer 11

so i just need to write in my notes when you convert to polar, r->0?

Answer 12

yes, thats what you're using to evaluate the limit

Answer 13

but r would not goto 0, if i had say (x,y)-> (2,3)?

Answer 14

i wonder if in that case we just wouldnt use polar to evaluate it

Answer 15

or maybe r-> hypotenuse?

Answer 16

\(x=r\cos\theta\)
\(y=r\sin\theta\)

\(r^2=x^2+y^2\)


as \((x,y)\to(2,3)\), we get \(r^2\to 2^2+3^2=13 \)