Could you explain another way why x is set to y^2? The explanation given doesn't really explain it. She just says, hey, if I let x = y^2, we'll get y^4 in the denominator; however, I don't see that as a reason. Unless, this rule applies to all similar problems, simply set x = y^2, we can throw comprehension out the window as to why x is set to y^2. And why is it that you couldn't just plug in 0 for all x's instead of all that factoring; you would have had zero as a result as well. However, this is a much better explanation than an online tutor with a bachelors degree who tried to wiggle out, not wanting to admit he does not know.
In my opinion, she choose the path x=y^2 because this path through the point(0,0) and another path is same. In class, I was taught to use another way. You can see that the exponent of y is double of x. So we take x=ky^2, k is constant, and substitute it in the limit. We obtain the constant contain k. You choose two value of k and get that the value of limit doesn't same. So, conclusion the limit doesn't not exist. There are many way to prove the limit doesn't not exist such as Hospital's rule, Sandwich theorem,... you can choose the best way to solve.
What happens if the coordinate points are not the same? ie (0,1) instead of (0,0)? I've looked over several different videos explaining the two-path limit test and literally every single one uses the point (0,0) -_-
Then you try another path until you get a different value. The point of the two-path test is to prove a limit does not exist, so you test different paths until you get a different value. I am not sure what it means if you get the same value for EVERY path you test, but I think it means the limit does exist.
wow, this is amazing. this helped me after a night of struggling. countless thank you's and gratitude is not enough. you are AMAZING!!!
thaks for your work,
it has been useful to me,
I want you to know that I am a Mexican student, so your help goes beyond borders, thank you very much
I have my multivariable calc midterm tomorrow, and this helped a lot! thank you!
howd you end up doing?
@@fernandor3136 i got a 98 hehe
@@sarahbauer2537 no way
Thank you ma'am as your teaching method is really cool to grasp everyone ...
thank you so much catherine for being great teacher.
very helpful explanation thank you merci beaucoup
Thanks you so much for making it far more comprehensive
i wish you were my teacher 😍
Could you explain another way why x is set to y^2? The explanation given doesn't really explain it. She just says, hey, if I let x = y^2, we'll get y^4 in the denominator; however, I don't see that as a reason. Unless, this rule applies to all similar problems, simply set x = y^2, we can throw comprehension out the window as to why x is set to y^2. And why is it that you couldn't just plug in 0 for all x's instead of all that factoring; you would have had zero as a result as well. However, this is a much better explanation than an online tutor with a bachelors degree who tried to wiggle out, not wanting to admit he does not know.
In my opinion, she choose the path x=y^2 because this path through the point(0,0) and another path is same. In class, I was taught to use another way. You can see that the exponent of y is double of x. So we take x=ky^2, k is constant, and substitute it in the limit. We obtain the constant contain k. You choose two value of k and get that the value of limit doesn't same. So, conclusion the limit doesn't not exist. There are many way to prove the limit doesn't not exist such as Hospital's rule, Sandwich theorem,... you can choose the best way to solve.
This video has saved me, thank you so much
What happens if the coordinate points are not the same? ie (0,1) instead of (0,0)? I've looked over several different videos explaining the two-path limit test and literally every single one uses the point (0,0) -_-
Because they choose the origin
Thanks for all your excellent explanation
Why didn’t you just let y=0 or x=0 as one of your paths? Would have simplified everything
wouldnt that get you 0/0 when you calculate it. im confused
amazing still usefull in 2021 thank you madam
thanks madame
Ohhh thanks for you❤❤❤❤❤❤❤❤
can you choose your x or y to be equal to something twice?
Best explained
Thankuu so much ,,it was so helpful
Great video
This is beautiful.
Thank you! It help me a lot!
Thank you for teaching me
Well explained!
how do you know which path to choose?
can we let here path 1st x=y^2 and path 2nd y=x^2
But what if you wanted to make y=sqrt(x)??? That would leave you w/ both limits being 1/2???
Then you try another path until you get a different value. The point of the two-path test is to prove a limit does not exist, so you test different paths until you get a different value. I am not sure what it means if you get the same value for EVERY path you test, but I think it means the limit does exist.
Sir. F(x,y) =x+y is contiues or not..
Thanks 👍
upon me agr x^2+y^6 to kese hoga
Sir jaha pr yke power 4 hai waha pr y ke power 2 karne pr kaise solve hoga same question please sir ans bate dijiye
Thanks a lot ♥️
F(x,y) = xy/x2+y2 at (0,0) solve it
She teaches pretty well😊😊