Posted by Bryant on Tuesday, February 28, 2012 at 2:30am.
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?

calculus  drwls, Tuesday, February 28, 2012 at 2:59am
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.

calculus  Bryant, Tuesday, February 28, 2012 at 3:29am
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?

calculus  Bryant, Tuesday, February 28, 2012 at 3:42am
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.

calculus  drwls, Tuesday, February 28, 2012 at 8:44am
<<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.

calculus  Bryant, Tuesday, February 28, 2012 at 12:37pm
then what can i approach by as well cause i also tried y=x1 and that was just a mess.
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