Limits of Multivariable Functions

In summary, for equations b and c, the limit does not exist. However, for equation a, the limit exists and is (0,0).
  • #1
sdobbers
11
0

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?
 
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  • #2
You need to show it for all paths even y=x^2 or even y=sin(x).
 
  • #3
For all three problems?
 
  • #4
sdobbers said:

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 [itex]\theta[/itex] then that is the limit of the whole function. If that limit does depend on [itex]\theta[/itex] then the limit does not exist. Leave things like cos(y) and ey in terms of y and hope you can decide without changing that!
 
  • #5
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, [tex]\theta[/tex]) --> (0,0)

(rcos[tex]\theta[/tex]rsin[tex]\theta[/tex]cosy) / ((r^2)(3(cos([tex]\theta[/tex])^2) + sin([tex]\theta[/tex])^2))

which gave me

(cos[tex]\theta[/tex]sin[tex]\theta[/tex]cosy) / (3(cos([tex]\theta[/tex])^2) + sin([tex]\theta[/tex])^2)

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


b) (rcos[tex]\theta[/tex]rsin[tex]\theta[/tex]) / r = cos[tex]\theta[/tex]sin[tex]\theta[/tex] limits goes to zero?

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

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

then I reduced it to

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

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

FAQ: Limits of Multivariable Functions

What are multivariable functions?

Multivariable functions are mathematical functions that have more than one independent variable. This means that the output of the function is dependent on multiple input values.

What are the limits of multivariable functions?

The limits of multivariable functions refer to the values that the function approaches as the independent variables get closer and closer to a certain point. It can also refer to the maximum and minimum values that the function can reach.

How do you calculate limits of multivariable functions?

To calculate the limits of multivariable functions, you can use the same principles as calculating limits for single-variable functions. You evaluate the function at different points approaching the desired point and see if the values approach a specific number.

What is the importance of understanding the limits of multivariable functions?

Understanding the limits of multivariable functions can help in determining the behavior and characteristics of the function. It can also be useful in applications such as optimization problems and understanding the shape of a surface.

Can multivariable functions have more than one limit?

Yes, multivariable functions can have multiple limits as the input variables can approach different values from different directions. This can result in different limiting values, or the function may not have a limit at all.

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