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Lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)

  • #1

Homework Statement


lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)


Homework Equations





The Attempt at a Solution


I have attempted to solve this by letting y=x giving me

lim (x^2*sin^2(x))/(x^2+2x^2) = lim x^2sin^2(x)/3x^2 = sin^2(x)/3 = 0
(x,x)->(0,0) (x,x)->(0,0)

And by letting both x and y = 0, both of which gave me the limit as being equal to 0 (left out equation, but can add in case I'm doing something wrong). The problem is that while all of these have given me a limit equal to zero, I don't know when I'm done and can say that the limit equals zero. We haven't gone over substituting in polar coordinates and when I tried to see how to do that I ended up confusing myself. Any easy to follow steps? (I did look back at other topics with the same problems, but wasn't able to work it out)
 

Answers and Replies

  • #2
61
0
Technically speaking, you're never done. That's because you have to prove that the limit exists approaching 0 for every path. We obviously can't prove EVERY path, because there's an infinite amount of paths approaching (0,0) from every direction. So we basically accept that the limit exists once we've proven that it exists from an acceptable number of paths... there really is no particular number of paths to prove...

Some other paths you might want to try:
y= (-x)
y= αx (if the limit you find has α in it, then it relies on α and the limit would then not exist)
y= x2

Cylindrical:
(just substitute)
let x=rcos∅ and y=rsin∅
 
  • #3
OK, so if I prove that the limit exist using, say, 5 or 6 different paths and they all come out to be zero, then I would be able to say that the limit exist, knowing full well that we haven't gone over every possible pathway? I keep reading that polar coordinates are supposed to be helpful with this topic, but he refuses to teach it and doesn't want us to use them if we know them. (Don't ask, it's a long story)

Sorry, I'm just trying to understand. It seems more like we're trying to prove the limit doesn't exist more then anything else.
 
  • #4
61
0
That's exactly it. It's much easier to prove that it doesn't exist than to prove that it does. Say you prove it doesn't exist just once... well then, it doesn't exist.

And yes, you could show it exists for 5 or 6 paths, knowing that you haven't tested them all... and the general consensus would be that it does exist... but some genius may just come along and prove you wrong by some obscure path.
 
  • #5
61
0
This guy has some good examples and explanations on multivariable limits:
 
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  • #6
Ray Vickson
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Homework Statement


lim (x,y)->(0,0) (x^2*sin^2(y))/(x^2+2y^2)


Homework Equations





The Attempt at a Solution


I have attempted to solve this by letting y=x giving me

lim (x^2*sin^2(x))/(x^2+2x^2) = lim x^2sin^2(x)/3x^2 = sin^2(x)/3 = 0
(x,x)->(0,0) (x,x)->(0,0)

And by letting both x and y = 0, both of which gave me the limit as being equal to 0 (left out equation, but can add in case I'm doing something wrong). The problem is that while all of these have given me a limit equal to zero, I don't know when I'm done and can say that the limit equals zero. We haven't gone over substituting in polar coordinates and when I tried to see how to do that I ended up confusing myself. Any easy to follow steps? (I did look back at other topics with the same problems, but wasn't able to work it out)
If I were doing it I would first note that as y -->0 we can replace sin(y) by y in your function F(x,y), so I would look at F1(x,y) = x^2 * y^2 /(x^2 + 2y^2)., which is >= 0 for any (x,y) not (0,0). Next I would note that if I have a smaller denominator I would have a larger fraction, so I would look at F2(x,y) = x^2 * y^2/(x^2 + y^2), because we have 0 <= F1(x,y) <= F2(x,y). Then in F2 I would switch to polar coordinates.

RGV
 
  • #7
Thanks a lot. I think I get it now, and the video did help some. I am going to work on using polar coordinates, even if we aren't allowed to use them. Seems to be handy even I use it just to check. Thanks again.
 

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