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Multivariable continuity using limits

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Capture.PNG

    2. Relevant equations
    lim(x,y)->(a,b)f(x,y) continuous at (a,b) if lim(x,y)->(a,b)f(x,y)=f(a,b)
    Squeeze theorem if lim a=lim c and lim a<= lim b <= lim c then lim b= lim c

    3. The attempt at a solution

    I proved that all the limits exist but somewhat the functions aren't all continuous. I don't know what I did wrong but the answer I submitted (all continuous) is wrong.

    1) using the squeeze theorem on the absolute value of g(x,y) i get 0<= 8x^2y^2/(x^2+y^2) <=8y^2 (since x^2/(x^2+y^2)<1) so the limit at (0,0) is equal to 0, which is also equal to f(0,0), hence the function is continuous

    2)I used the squeeze theorem once again and replaced x^3/(x^2+y^2) and am left with limit of 7xsiny which gives 0 so it is continuous

    3) I expressed the limit as a substration of two limits, and using the squeeze theorem on each of them got 0. I once again used x^2/(x^2+y^2)<1 and y^2/(x^2+y^2)<1 to get limit of xy - limit of 9y^2 which both equal to 0.

    4) using the squeeze theorem and the fact that x^2/(x^2+y^2)<1, I get limit of 6y which equals 0

    5) Same as 3

    In theory I proved that all the limits exist and are equal to 0 so all the function should be continuous. Does anyone have any clue regarding what I did wrong? Thanks!
     
  2. jcsd
  3. Nov 2, 2015 #2

    LCKurtz

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    So you have proven that ##f(x,y)## is continuous at ##(0,0)##. But the problem asks whether it is continuous in the whole plane. For example, look at case 5. What happens if ##x=1## and ##y\to 0## so you are approaching ##(1,0)##? Do you get ##f(1,0)##?
     
  4. Nov 2, 2015 #3
    In case 5, f(1,0) is equal to 0. The limit if we set x=1 will become lim y->0 [6(1)^2y/(1^2+y^2)]=lim y->0 [6y/1+y]=0, so I get the same value.
    We need to prove that the function is continuous on the whole plane, however we already know that the given functions are rational functions, so they are continuous at all points except when the denominator is equal to 0. In all the rational functions given, the discontinuity occurs at (0,0), but since we are given a specific value, namely 0, for this point on the plance, we only need to prove continuity at this point. Any input? I think I redid the calculations for 5 correctly but I might be wrong. Thanks!
     
  5. Nov 2, 2015 #4

    LCKurtz

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    That wasn't case 5.
     
  6. Nov 2, 2015 #5
    Oh I see I was only checking at (0,0) not for (xi, 0) (0, yi). Thanks!
     
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