Question

[CALCULUS III—CONTINUITY] Discuss the continuity of function f(x, y). figure inside

Answer 1

\[f(x,y)=\frac{xy^2}{x^2+y^2}\]

Answer 2

I am inclined to say the function is continuous in R3 because there are no real numbers that can produce a faulty denominator.

This is purely thought, I have not investigated any further in regards to limits or such.

Can anybody verify the continuity?

Answer 3

graph of f(x, y) http://www.wolframalpha.com/input/?i=f%28x%2Cy%29%3Dxy^2%2F%28x^2%2By^2%29

Answer 4

@calmat01 continuity of f(x, y)?

Answer 5

(see attachment)

wolfram describes the domain as \[(x, y) \in R^2 : x^2 + y^2>0\]

Does this imply continuity at all values x, y? What confuses me is the : x^2 + y^2 > 0

Answer 6

limits — http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

Answer 7

definition of continuity (see attachment)

Answer 8

Hmm I always hated these problems. Because if it IS continuous it feels like you just have to keep looking for possible discontinuities until you find one (which you won't).

I think here \((x, y) \in R^2 : x^2 + y^2>0\) they're just saying, The Domain is the set of all points (x,y) within R2 such that ... hmm I guess they're describing the region as a circle greater than zero.

Could there maybe be a discontinuity at zero? :o hmm

Answer 9

@zepdrix Possibly. Using the definition of a limit and continuity, let us evaluate the lim x, y -> 0

Answer 10

/rightarrow

is a good keyword to remember :) heh

Answer 11

\rightarrow*

Answer 12

\[\lim_{(x,y)\rightarrow0}\frac{xy^2}{x^2+y^2}\]

Answer 13

my reference example for this approach: http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx#PD_Limit_Ex1d

Answer 14

y = x\[\lim \frac{x(x)^2}{x^2+(x)^2}=\lim x^3/2x^2=\lim x/2=0\]

Answer 15

I think what they mean is that no matter what value of x or y you choose, the result will always be a number greater than zero, as long as x nor y are zero...lol

Answer 16

And, again, using the reference, I am going to try \[y=x^3\]\[\lim\frac{x(x^3)^2}{x^2+(x^3)^2}=\lim x^7/(x^2+x^6)=\lim \frac{x^5}{1+x^4}\]

Does this turn into L'ohpital's where the lim is ≠ 0? Therefore the function is discontinuous at the point?

Answer 17

@calmat01 That is possibly. Does that mean I am evaluating the limits incorrectly? What you say is an approach I have yet to consider

Answer 18

Oh jees, whoops, the lim = 0

Answer 19

So there is no consideration of Lohpitals rule, the limit checks out both considerations to be 0. So I am wanting to go with calmat01's explanation of the inequality provided by wolfram.

Answer 20

Do the three of us conclude that the function is continuous at all points?

Answer 21

Well, we know we cannot divide by zero, and the only way x^2+y^2 to be zero would be if both x and y were zero. Otherwise, you get a nonzero number.

Answer 22

Hmm you can approach zero from the axis.
Example: y=0
\[\large \lim_{(x,0)\rightarrow(0,0)}\frac{0}{x^2}\]Hmm that doesn't seem to help us though.

Answer 23

I say continuous :3 I dunno lol

Answer 24

I would say continuous provided neither x nor y are zero...that's a cop out, I know.

Answer 25

So did we come to a consensus?

Answer 26

Seems good. we will go with continuous.

Answer 27

the only point of discontinuity is at (0,0) because \[x^2+y^2=0\] you cant have 0 in the denominator. It's a point of removable discontinuity.