Limits and Continuity Using Polar Forms

0:00 Introduction
0:18 Lim [(x^3+y^3)/(x^2+y^2)] as (x, y) approches (0,0).
3:19 Lim [(x^2+y^2) ln(x^2+y^2)] as (x, y) approches (0,0). Using L'Hospital Rule.
7:15 Lim [(e^(-x^2-y^2)-1)/(x^2+y^2)] as (x, y) approches (0,0). Using L'Hospital Rule.
Thank you for watching!
JaberTime

[@user-jq9qs8pp4u]

At the third question, how can I induce it without using lhopital's rule?

[@19917119]

Hi, Since with direct substitution we end up with the form " 0/0", we will have to use L'Hopiyal's Rule.
Here is a link that I use as x approaches 0 for the same problem using an online software:
www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto%200%7D%5Cleft(%5Cfrac%7Be%5E%7B-r%5E2%7D-1%7D%7Br%5E%7B2%7D%7D%5Cright)?or=input
JaberTime

[@user-jq9qs8pp4u]

@@19917119 😊 thank you

[@guiolaio1514]

@@19917119 Not necessarily, we could just multiply and divide by -1, and then apply the substitution t = -r^2, and we would get - lim t->0 e^t -1 /t , and this is a standard limit, which would yield 1 in itself, so the final result would be -1

[@tibrummichael2715]

add the videos that involve sin and cos

[@19917119]

Hi, The polar Forms do uses the sine and cosine. However, I will look for more examples that involve sin and cos.