Limits of Multivariable Functions

[sdobbers]

Homework Statement

Find the limit, if it exists, or show that the limit does not exist.

limit (x,y) --> (0,0)

a) f(x,y) = (xycosy) / (3x^2 + y^2)

b) f(x,y) = (xy) / sqrt(x^2 + y^2)

c) f(x,y) = ((x^2)ye^y) / ((x^4) + 4y^2)

Homework Equations

The Attempt at a Solution

a) For this one, I did (0,y)-->(0,0) and got 0; then did (x,0) --> (0,0) and got 0. Then I substituted y=x, so (x,x) --> (0,0). I ended up getting ((x^2)cosx) / 4x^2; which I evaluated as x goes to zero, the limit would not exist (since the bottom would be zero).

b) Again I did x=0, and y=0 and came up with 0 for both of those. Would I then, substituted y=x again? Ending up with x^2 / xsqrt(2), so x/sqrt(2) goes to zero as x goes to zero. So the limit would be zero. Would I have to try more paths?

c) Again, x=0, y=0 resulted in limit of 0. Next I attempted x = sqrt(y); so ((y^2)e^y) / ((y^2) + 4y^2) which lead to (1/5)e^y. So as y goes to zero, e^y goes to 1, and the limit goes to 1/5. Therefore, the limit does not exist.

Did I go about these in the right way, and would I need to test more paths for each of them?

 

[JasonRox]

You need to show it for all paths even y=x^2 or even y=sin(x).

 

[sdobbers]

For all three problems?

 

[HallsofIvy]

Homework Statement

Find the limit, if it exists, or show that the limit does not exist.

limit (x,y) --> (0,0)

a) f(x,y) = (xycosy) / (3x^2 + y^2)

b) f(x,y) = (xy) / sqrt(x^2 + y^2)

c) f(x,y) = ((x^2)ye^y) / ((x^4) + 4y^2)

Homework Equations

The Attempt at a Solution

a) For this one, I did (0,y)-->(0,0) and got 0; then did (x,0) --> (0,0) and got 0. Then I substituted y=x, so (x,x) --> (0,0). I ended up getting ((x^2)cosx) / 4x^2; which I evaluated as x goes to zero, the limit would not exist (since the bottom would be zero).

b) Again I did x=0, and y=0 and came up with 0 for both of those. Would I then, substituted y=x again? Ending up with x^2 / xsqrt(2), so x/sqrt(2) goes to zero as x goes to zero. So the limit would be zero. Would I have to try more paths?

c) Again, x=0, y=0 resulted in limit of 0. Next I attempted x = sqrt(y); so ((y^2)e^y) / ((y^2) + 4y^2) which lead to (1/5)e^y. So as y goes to zero, e^y goes to 1, and the limit goes to 1/5. Therefore, the limit does not exist.

Did I go about these in the right way, and would I need to test more paths for each of them?

Showing that you get the same limit for some paths is not enough- there still might be some, perhaps very complex, path for which you would get a different result. JasonRox is correct that you would have to show that you get the same limit for every possible path but, in general, there is no good way to do that.

I would recommend changing into polar coordinates. That way the distance from (0,0) is measured only by the variable r. If the limit, as r goes to 0, does not depend on \theta then that is the limit of the whole function. If that limit does depend on \theta then the limit does not exist. Leave things like cos(y) and e^y in terms of y and hope you can decide without changing that!

 

[sdobbers]

I've never really been any good at converting from Cartesian to Polar (for some reason my college just doesn't seem to do it that often). So I'm not really sure if I'm doing this right.

a) converted the first equation into (r, \theta) --> (0,0)

(rcos\thetarsin\thetacosy) / ((r^2)(3(cos(\theta)^2) + sin(\theta)^2))

which gave me

(cos\thetasin\thetacosy) / (3(cos(\theta)^2) + sin(\theta)^2)

From here, where do I go? Or would I just say that since the r's canceled it's only dependent upon \theta, which means it does not exist?


b) (rcos\thetarsin\theta) / r = cos\thetasin\theta limits goes to zero?

c) This one I have no clue where to go, here's what I attempted, I converted it to (r, \theta):

((r^2)(cos(\theta)^2)rsin\theta(e^y)) / ((r^4)(cos(\theta)^4) + 4(r^2)sin(\theta)^2)

then I reduced it to

(r(cos(\theta)^2)sin(\theta)(e^y)) / ((r^2)(cos(\theta)^4) + 4(sin(\theta)^2))

Then I get stuck, what do I do from there? =/