Question

Limits question below!

Answer 1

I got you :) hold on.

Answer 2

What have you tried? Have you checked what happens along different paths?

Answer 3

Okay I have it figured out now.

Answer 4

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto0%7D%5Cleft(%5Cfrac%7Bxy%2Byz%5E%7B2%7D%2Bxz%5E%7B2%7D%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%2Bz%5E%7B4%7D%7D%5Cright)

Answer 5

https://www.symbolab.com/
This is an amazing calculator :) It literally does everything :)

Answer 6

That calculator doesn't seem to support limits in several variables, so that's not right. Plus that result is hardly useful. There are infinitely more ways to approach the origin than by just letting \(x\to0\).

Answer 7

well I helped her with one part if you would like to continue to help her then go ahead and do so. she can enter more than just 0 that's why I put down the link so she can continue on.

Answer 8

If I plug in the variables, it's clearly "0/0". I was wondering if there was another way I can go around it to prove that it either DNE or exists.

Answer 9

Perhaps L'Hopital's?

Answer 10

That's just the thing; what you posted doesn't really help with anything.

One easy way to compute the limit is to just set one of the variables equal to \(0\), namely \(z=0\), so that we confine any path we choose to the \(x,y\) plane. That takes care of \(z\to0\).

So you're left with examining the limit of a bivariate function:
\[\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+y^2}\]

Answer 11

This is much easier to work with. I suggest examining what happens when you choose \(y\) as any line through the origin.

Answer 12

Holster seems to know more so i'm gonna let him help you :) I hope you get this right :)

Answer 13

Well... Let's see.
I can let x=0, then the limit doesn't exist.
Then I let y=x (as you've suggested), then the limit is x^2/(2x^2)=1/2
So that means the limit doesn't exist since the two limits don't match up?

Answer 14

Also, would converting them to polar coordinates be of any significance?

Answer 15

Exactly right. More generally, if you fixed \(y=mx\) (an arbitrary line of slope \(m\) through the origin), then you're left with
\[\lim_{x\to0}\frac{mx^2}{x^2+m^2x^2}=\frac{m}{1+m^2}\]so the limit depends on \(m\).

Not sure about cooordinate conversion. Nothing immediately convenient comes to mind, though it's possible you could use cylindrical/spherical to your advantage.

Answer 16

@HolsterEmission Awesome! This helps a lot, thank you :)

Answer 17

yw, glad I could help!