Question

Go to this site: http://tutorial.math.lamar.edu/Classes/CalcI/ComputingLimits.aspx
Check out example c. In his solution he checks the limit with 3 different approaches. (x,0),(0,y),(x,x), when he evaluates the (x,0) and (0,y) he gets 0/0 but still evaluates it to 0 in the limit where it should be undefined unless (obviously I am) missing something. What am I missing?

Answer 1

I don't see an example "c"; I see examples 1 through 6

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx
Example 1.c, sorry, wrong link!

Answer 3

I think he's right, but leaves something out of the explanation.
When y=0 he has this limit: \[\lim_{x\rightarrow0} '{0 \over x^4}\]
When x=0 he has this limit: \[\lim_{y\rightarrow0} '{0 \over 3y^4}\]
Both of those are constant functions with a single removable singularity (it's technically not a discontinuity because that point is not in the domain). Since it's a constant function, the limit aproaching that point has the same value that the function has at every other point.

Answer 4

Ok, so how would you say ((x^2)*y)/(x^4+y^2) and ((x^2)*y)/(x^2+y^2) would evaluate to? I'm quite sure the first one is not defined in the point (0,0) but the second one is. What's your approach?

Answer 5

Neither is defined at (0,0), both hit division by zero.
I assume you mean how would I evaluate the limits along one axis as the other axis approaches zero, and it's the same as in the example:
\[\lim_{(x,0)\rightarrow(0,0)}'{x^2 \cdot y \over x^4+y^2} = \lim_{x \rightarrow 0}'{0 \over x^4} = 0\] \[\lim_{(0,y)\rightarrow(0,0)}'{x^2 \cdot y \over x^4+y^2} = \lim_{y \rightarrow 0}'{0 \over y^2} = 0\] \[\lim_{(x,0)\rightarrow(0,0)}'{x^2 \cdot y \over x^2+y^2} = \lim_{x \rightarrow 0}'{0 \over x^2} = 0\] \[\lim_{(0,y)\rightarrow(0,0)}'{x^2 \cdot y \over x^2+y^2} = \lim_{y \rightarrow 0}'{0 \over y^2} = 0\] As in the example, this isn't enough to show that either limit actually exists, it just fails to show that either limit does not exist.

Answer 6

So, what would one how to do to actually see if either if the limit does exist or does not exist?

Answer 7

As far as I know there is no sure way to prove that the limit does exist; if there is, I doubt it falls within the scope of Calc 3.
A quick search turned up http://bit.ly/glqw77 , which may give a sure way; I don't have time to confirm that right now.

Answer 8

Thanks, seems a bit weird that it's so hard to define/solve really.