= lim r->0 r^4(cos^4(theta)+sin^4(theta))/r^2 = r^2(cos^4(theta)+sin^4(theta))=0
that is a standard limit, it is probably somewhere in your textbook.
write it like ln(r^2)/(1/r^2) and apply L'hopital to see that is equal to lim r->0 -r^2
So, help on any or all questions would be appreciated. I thought I knew what I was doing, but these threw me for a loop. So, I ask you, dear stranger, teach me to swim. I keep sinking!
1. Use polar coordinates to find the following limits. Clearly show steps and explain your reasoning as necessary.
a.) lim (x,y) -> (0,0) (x^4+y^4)/(x^2+y^2)
I simplified this to get an answer of zero. Is this correct. Btw, relevant equations x=rcos(theta) y=rsin(theta) r^2=X^2+y^2
b.) lim (x,y) -> (0,0) (x^2+y^2+1) ^ (1/(x^2+y^2))
Of course, this is equal to lim r -> 0 (r+1) ^ (1/r). I thought this would head to 1, but the answer is e. I have no idea how!
c.) lim (x,y) -> (0,0) (X^2 + Y^2) ln (x^2+y^2)
Again, simplified to lim r->0 r^2 ln (r^2)
I have no idea how to do this. I thought since r is going to zero, the whole thing would. But, that doesn't seem right for some reason.
2.) IF z^3-xz-y=0 prove that (partial^2 z)/(partial x partial y) = - (3z^2 +x)/(3z^2 -x)^3
Alright, for this I changed the first part (the partial nonsense) to partial with respect to x of ( - partial f/ partial y)/(partial f/partial z)
where f = z^3 -xz-y
partial f/partial y = -1
partial f/partial z = 3z^2-x
which means the whole partial nonsense equals
but when i work it all out, i get a negative on one side but not the other.
Thank you in advance! =)