Okay, so apparently the limit does NOT exist, although I have yet to find a contradiction, that is, a path that does not equal 0 .
Hello.
Determine the following limit, or show it doesn't exist.
lim [(x,y) -> (1,0)] [2(x - 1)*y^(3/2)]/[x^2 - 2x + 1 + y^3]
WORK:
So I tried several paths.
Along x = 1, we have:
lim [(1,y) -> (1,0)] [2(0)y^(3/2)]/y^3 = 0/y^3 = 0
Along y = 0, we have:
lim [(x,0) -> (1,0)] 0/[x^2 - 2x + 1] = 0
So, it *appears* the limit exists so far, but in reality it exists until you can prove otherwise, that is, find a path that does not = 0. If it does exist, how do I show that it exists (and there is, in fact, NOT a path out there that does not = 0).
Okay I see what the problem is.
lim (x,y)-->(1,0) of [2(x-1)y^(3/2)]/[(x-1)^2+y^3]
You can simplify this as,
lim (x,y)-->(0,0) of [2xy^(3/2)]/[x^2+y^3]
Because, x-1 approaches x as x approaches 1.
So you can replace the limit coordinates with this substitution.
And now, try different paths.
Thanks for the help, TPH. I never would have thought of doing this.. quite clever!
However, I tried many paths and I still get the limit is 0..
WORK:
So we have lim (x,y)-->(0,0) of [2xy^(3/2)]/[x^2+y^3]
Let x = y; then, lim (y,y) -->(0,0) of [2y*y^(3/2)]/[y^2 + y^3] = lim (y --> 0) of [2y^(5/2)]/[y^3 + y^2] = lim (y --> 0) of [2*sqrt(y)]/[y + 1] = 0..
Similarly, if you let y = x you'll get the same thing.
If you let x = 0, you get
lim (0,y) --> (0,0) of [0/y^3] = 0
And if you let y =0, you get the same thing...
Still can't find a path where the limit is not 0.