## anonymous 5 years ago 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?

1. anonymous

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

2. anonymous

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

3. anonymous

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.

4. anonymous

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?

5. anonymous

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.

6. anonymous

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

7. anonymous

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.

8. anonymous

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