lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^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^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
= Lim (x^3-1)/(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=x-1 and that was just a mess.
To find the limit as (x, y) approaches (1, 0) of the expression (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2), you can substitute the values x = 1 and y = 0 into the expression and simplify.
So, let's evaluate the expression at (x, y) = (1, 0):
lim (x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)
(x,y)->(1,0)
Substituting x = 1 and y = 0, we have:
lim (1^3-1)/(1^3+2(1^2)(0)+(1)(0^2)-1^2-2(1)(0)-0^2)
Simplifying further:
lim (0)/(1+0+0-1-0-0)
lim 0/0
We have encountered an indeterminate form of 0/0, which means we cannot determine the limit directly from this evaluation.
To see if the limit changes when approaching (1, 0) along the line y = x, we can substitute y = x into the original expression and evaluate the limit at (x, x) = (1, 1):
lim ((x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)) when y = x
(x, x) -> (1, 1)
Substituting y = x and evaluating at (x, x) = (1, 1):
lim ((x^3-1)/(x^3+2x^2y+xy^2-x^2-2xy-y^2)) when y = x
(x, x) -> (1, 1)
lim ((x^3-1)/(x^3+2x^2x+x^3-x^2-2x^2-x^2)) when y = x
(x, x) -> (1, 1)
lim ((x^3-1)/(3x^3-4x^2)) when y = x
(x, x) -> (1, 1)
Substituting x = 1:
lim ((1^3-1)/(3(1)^3-4(1)^2)) when y = x
(1, 1) -> (1, 1)
lim (0)/(3-4)
lim 0/-1
In this case, the limit evaluates to 0/-1, which is equal to 0.
Therefore, based on the evaluation, the limit as (x, y) approaches (1, 0) is indeterminate, but the limit as (x, x) approaches (1, 1) is 0.