mathematics function limits

Question

a) lim(x,y) → (1,1) lim 2x³+2y / y²x+1 b) lim(x,y) →(1,0) x²y²-y² / x-1 c) lim(x,y) → (0,0) 3xy-y² / x²+y²

Answer 1

a)

This function is continuous for (x,y)→(1,1) so we can just substitute in (1,1)

lim = 4/2 = 2

If it was (-1,1), then we would have issues because there would be a discontinuity.

b) I looks like we have a discontinuity at x=1, but this can be removed just like you do with one variable.

(x²y²-y²)/(x-1) = y²(x²-1)/(x-1) = y²(x+1)(x-1)/(x-1) = y²(x+1)

lim = 0(0+1) = 0

c) Obviously there is a discontinuity at (x,y) = (0,0). So, we need to take more limits, such as

(x,0)→(0,0) and (0,y)→(0,0). If we come up with different answers, then the limit diverges and the

limit does not exist (DNE). If we get the same answer, we have to continue with x=y, y=x², etc. until we

get different answers or are satisfied that we have a limit.

lim (x,0)→(0,0) = (3(x)(0)-0)/(x²+0) = 0

lim (0,y)->(0,0) = (3(0)y-y²)/(0+y²) = -y²/y² = -1

We have two different answers, so the limit DNE.