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Limit with 2 variables

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that the following limit exists and find its value.

    limit(x,y approaches 0,0) [ (y^3) / (|x| + y^2) ]


    3. The attempt at a solution
    this was on a test and this is what i did. i got 4/12 pts on this.

    i did the limit along the x-axis (y=0)
    :
    f(x,0) = (0^3) / ( |x| + 0 ) = 0
    so the limit = 0 along x axis

    along y axis (x=0):
    f(0,y) = (y^3) / (0 + y^2) = y
    and the limit (y goes to 0) y = 0

    along x = y^2:
    f(y^2,y) = (y^3) / (y^2 + y^2) = (y^2*(y)) / (y^2(1+1)) and y^2 cancels leaving y/2
    and the limit y goes to 0 of y/2 = 0 along x=y^2

    therefore limit(x,y approaches 0,0) [ (y^3) / (|x| + y^2) ]
     
  2. jcsd
  3. Nov 4, 2011 #2

    LCKurtz

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    All you have shown is that the limit is zero along a few paths. But that doesn't show the limit is 0 along all paths. You can never show a limit exists with that technique. You need to use a δ-ε technique and show the fraction can be made small as (x,y)→(0,0).

    I would suggest changing it to polar coordinates to get started and think about estimating things as r → 0.
     
  4. Nov 4, 2011 #3
    But that's what the book shows and what we've done in class and for home works ..
     
  5. Nov 4, 2011 #4

    LCKurtz

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    No it isn't. (I wasn't there but I'm pretty sure about that.) If you are working a problem where the limit doesn't exist, if you can find a couple of paths where the limits are different, that would prove the limit doesn't exist. The reason is that if the limit does exist, all paths would give the same answer.

    But you can not prove a limit does exist by checking a few paths.
     
  6. Nov 4, 2011 #5
    well, i dont really see the problem as asking IF it exists... it just says to show that it does. so i mean its kinda already telling you..
     
  7. Nov 4, 2011 #6
    and, yes, that is indeed what we've been doing in class and in all recitations for these types of problems.
     
  8. Nov 5, 2011 #7

    LCKurtz

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    I still don't believe you. If that is how you are "proving" the limit exists, you need a new teacher.

    The point in the author telling you to show the limit exists is probably so you won't waste time trying to find paths that give different limits. It is telling you the answer: "The limit exists." But asking to you "show it exists" means you are to prove it, not just take the author's word for it.

    But, hey, if you don't want to do the problem, don't do it.
     
  9. Nov 5, 2011 #8
    @arl146: LCKurtz is correct. To use an equivalent example; Person 1 says, Prove God exists and Person 2 says to person 1, you can't prove God doesn't exist, so God exists. (Hopefully that's how the analogy is meant to go). Disproof and proof aren't the same. Be careful, maybe you misunderstood the teacher.
     
  10. Nov 5, 2011 #9
    The limit exists iff for every ε>0 there exists a δ>0 such that for all (x,y) with 0 <|(x,y)-(p,q)|<δ, then |f(x,y)-L|<ε. Try this and you should be able to prove it, it's a lot less work than checking the infinite amount of curves on which L exists.
     
  11. Nov 5, 2011 #10
    i went with a friend to a place on campus to get homework help and more than 1 TA told us to do it that way then use the Squeeze Theorem.
    it doesnt even matter! i still dont know how to do it!
     
  12. Nov 5, 2011 #11

    LCKurtz

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    You might try the suggestion above that I made in post #2.
     
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