i don't understand the "*. *". Comment. This site is not about PROVING someone is wrong or putting people down which you did with me. The dialog in these forums is not supposed to be judgmental towards others. So tell me why you disagree without being judgmental. What do you see in detail is...
How about this approach. I plotted this function using quick graph. There are no breaks in the graph at the origin so therefore it is continuous there. Also the numerator sin(x^3 + y^3) can be rewritten as sin(x^3)cos(y^3) + cos(x^3)sin(y^3). Now that the numerator has been rewritten as a sum...
That makes complete sense when using the squeeze theorem for the path y(x) =kx. Earlier in the posts where y=x was tried it, would have worked then if the squeeze theorem was applied at that point in time to the problem. I forgot all about suggesting that theorem be tried again. Thank you!
I am not saying I agree with the answer, just that I have never tried that approach before. One problem with letting x=rcos(t) and y=rsin(t) is that it limits curve to travel at any point on the unit circle (assuming r=1 for arguments sake) for any value of t, which supports your comment about...
I would say no because this path may not work for other limit problems. Under the reasoning that if there was a "family" of paths that satisfied must multi variable limit problems it would have brought up when I took calculus courses years ago. What are your thoughts?
I want to clarify my use of polar in terms of the equations that define a circle. I think you are describing a disk in the x,y plane approaching a radius of 0 or in three space a cone who radius is converging to 0
Interesting approach to do it from a polar equation point of view. Let me think about. Are we assuming a fixed value of t or is r and t variable. Based on what you wrote it looks like t is fixed. Maybe you can clarify that for me thanks.
This is not multiplication of f(x) by g(y). It is f(y)/g(y). l'hopitals rule only applies to single variable limits of functions. I am trying to get you visually see how each path in a multi variable limit looks like one variable at a time. You will need to be creative in your path selections...
I can see an operation on the function of f(x) or f(y) for the case of a indeterminate limit do you see it? And it works for g(x,y) = the sin function and h(x,y) =x+y^2. When I apply this rule my limit (x,y) -> (0,0) does equal 0. Could you explain why you want to prove this with the...
Let's say x is turned off then you have f(y) = sin(y^3)/y^2 ? The limit (0,y) ->(0,0) is indeterminate at 0/0 and the limit is not 0. Likewise for (x,0) ->(0,0). Can you solve the limit it as functions of one variable for the function sin(y^3)/y^2? Hope this helps
What do you get if you set x=0 and y be any value and take the limi? and do the same for y=0 and x any value and take the limit? Then try the limit of the function over the path x=y. In order for the limit to exist all of the selected paths must be continuous and have the same limit.