calculus
posted by Bryant on .
lim (x^31)/(x^3+2x^2y+xy^2x^22xyy^2)
(x,y)>(1,0)
my question is can you approach (1,0) with y=x and does that change the the limit to
lim f(x,x)
(x,x)>(1,1)
in which case i get 3/4 as the limit. and if this is not how i go about doing it can you point me in the right direction?

You cannot approach along y = x because that relationship does not apply at (1,0).
Try approaching along y = 0
Lim (x^31)/(x^3+2x^2y+xy^2x^22xyy^2)
= Lim (x^31)/(x^3 x^2)
x>1
= Lim 3x^2/(3x^2 2x)
x>1
= 3
I used L'Hopital's rule. 
alright thanks, my teacher never showed us that if you set one of the variables equal to zero you can make it a one variable function. so say i approach x=0 and find the limit of that to be infinity. does the original function f(x,y) not exist?

or does x=0 not work because x does not equal zero on the point (1,0). cause my teacher was fairly lazy in class and used the point (0,0) for every example.

<<or does x=0 not work because x does not equal zero on the point (1,0)? >>
It does not work because x is not 1 at y = 0, where they want the limit evaluated. 
then what can i approach by as well cause i also tried y=x1 and that was just a mess.