Thank you very much, your approach gives me a better understanding to solve multivariate limits. During last assignment, I only pick several paths (when the outcomes are the same) and said that is the limit instead of using squeeze theorem to prove it. Wish I viewed your video sooner. It helps a lot. Is it possible you can make more videos about MAT237? Appreciate that.
Thanks very much for your comment, I'm glad that the video has been helpful to you. I have some videos specific to MATH 237 that you can access in my MATH 237 playlist: th-cam.com/play/PLL9sh_0TjPuOxWvcGm5AOpxL6g3bmqnzb.html. These videos are from the spring 2021 iteration of the course. If there are topics that are not included in this playlist, you might find it helpful to review some of my other playlists on multivariable calculus. Remember though that your iteration of MATH 237 may present these topics differently, use different notation, or focus on other topics that are not covered in these videos. The videos on this channel are not officially affiliated with the course and should be used only as a supplemental resource.
Question: Why not switch to polar coordinates? We see that the denominator is x^2+y^2, which tells me this is a prime candidate and in polar, x^2+y^2 = r^2. x = rcos(theta), y = rsin(theta) (x^3+2y^3+2x^2+2y^2)/(x^2+y^2) becomes [(rcos(theta))^3 + 2(rsin(theta))^3 + 2(rcos(theta))^2 + 2(rsin(theta))^2] / r^2 We then simplify this to... [r^3 * cos^3(theta) + 2r^3 * sin^3(theta) + 2r^2 * cos^2(theta) + 2r^2 * sin^2(theta)] / r^2 Factor out an r^2 from the numerator, and then cancelling it out with the denominator, we get r * cos^3(theta) + 2r * sin^3(theta) + 2* cos^2(theta) + 2* sin^2(theta) Factor out an r from the first 2 terms, and recognizing that cos^2 + sin^2 = 1, we get... r(cos^3(theta)+2*sin^3(theta)) + 2*1 (x,y) --> (0,0) is the same as r--> 0, so if we plug in r = 0, we get the value of 2 for the expression, which is the limit you got as well. Simpler, no?
You explained the Squeeze Theorem better than my prof did! Thank you! The Triangle Inequality was something I never learned, it explains why I was confused at my university. So an equality becomes an inequality when applying the triangle inequality if I understood correctly? I just had trouble knowing when = becomes
If you're working with the absolute value of a sum of two (or more) terms, |a+b|, you can bring the absolute value into the sum at the cost of an inequality: |a+b|
love the enthusiasm man, you make these problems not scary at all!
Best approach I've seen for this kind of limits so far
Thanks Mathemation!
the heavy breathing at 7:21 🤣🤣🤣
I like the way you explained this. Very clear
Very helpful! Thank you!
very nice content, keep it up, thank you!
Thanks very much! I’m glad you’re enjoying the videos :)
Thank you very much, your approach gives me a better understanding to solve multivariate limits. During last assignment, I only pick several paths (when the outcomes are the same) and said that is the limit instead of using squeeze theorem to prove it. Wish I viewed your video sooner. It helps a lot. Is it possible you can make more videos about MAT237? Appreciate that.
Thanks very much for your comment, I'm glad that the video has been helpful to you. I have some videos specific to MATH 237 that you can access in my MATH 237 playlist: th-cam.com/play/PLL9sh_0TjPuOxWvcGm5AOpxL6g3bmqnzb.html. These videos are from the spring 2021 iteration of the course. If there are topics that are not included in this playlist, you might find it helpful to review some of my other playlists on multivariable calculus.
Remember though that your iteration of MATH 237 may present these topics differently, use different notation, or focus on other topics that are not covered in these videos. The videos on this channel are not officially affiliated with the course and should be used only as a supplemental resource.
@@mathemation Thank you for the playlist, very helpful. Yes, sir. I will focus on my textbook first and then learn what I missed from your video.
Question: Why not switch to polar coordinates? We see that the denominator is x^2+y^2, which tells me this is a prime candidate and in polar, x^2+y^2 = r^2.
x = rcos(theta), y = rsin(theta)
(x^3+2y^3+2x^2+2y^2)/(x^2+y^2) becomes
[(rcos(theta))^3 + 2(rsin(theta))^3 + 2(rcos(theta))^2 + 2(rsin(theta))^2] / r^2
We then simplify this to...
[r^3 * cos^3(theta) + 2r^3 * sin^3(theta) + 2r^2 * cos^2(theta) + 2r^2 * sin^2(theta)] / r^2
Factor out an r^2 from the numerator, and then cancelling it out with the denominator, we get
r * cos^3(theta) + 2r * sin^3(theta) + 2* cos^2(theta) + 2* sin^2(theta)
Factor out an r from the first 2 terms, and recognizing that cos^2 + sin^2 = 1, we get...
r(cos^3(theta)+2*sin^3(theta)) + 2*1
(x,y) --> (0,0) is the same as r--> 0, so if we plug in r = 0, we get the value of 2 for the expression, which is the limit you got as well.
Simpler, no?
You explained the Squeeze Theorem better than my prof did! Thank you! The Triangle Inequality was something I never learned, it explains why I was confused at my university. So an equality becomes an inequality when applying the triangle inequality if I understood correctly? I just had trouble knowing when = becomes
If you're working with the absolute value of a sum of two (or more) terms, |a+b|, you can bring the absolute value into the sum at the cost of an inequality:
|a+b|