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?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I don't see an example "c"; I see examples 1 through 6
http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx Example 1.c, sorry, wrong link!
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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?
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.
So, what would one how to do to actually see if either if the limit does exist or does not exist?
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.
Thanks, seems a bit weird that it's so hard to define/solve really.

Not the answer you are looking for?

Search for more explanations.

Ask your own question