Find the limit. lim as (x,y,z)-->(0,0,0) of (xy+yz+zx)/(x^2+y^2+z^2)?

Answer 1

the limit does not exist

choose two curves passing through the origin:

P1: x = y = z

limit becomes lim 3x^2 / 3x^2 = 1

P2: x = y/2 = z/3

limit becomes: lim (x*2x + 2x*3x + 3x*x) / (x^2 + 4x^2 + 9x^2)
= lim 11x^2 / 14x^2 = 11/14

§
thus the limit does not exist.

Answer 2

Christine: (x, y, z) -> (0, 0, 0) does not mean that we can replace y and z by x. This merely tests approaching the origin along one particular line; if the function behaves differently away from this line you won't get the correct answer, as in this case.

Along the line y=z=0:
0 / x^2 = 0

Along the line y=x, z=0:
x^2 / (2x^2) = 1/2

Since these are not equal, the limit does not exist.

Another easy line to check is x = y = z:
3x^2 / 3x^2 = 1.

For the limit to exist we must get the same limit along every possible trajectory. You should always check a few basic and easy-to-calculate trajectories first. If you can find two that don't give equal limits, you're done; if you can't, you can start looking at proving the limit does exist.

Note that there are functions on R^2 which have the same limit on all trajectories of form y = kx for constant k, but nevertheless do not have a limit at (0, 0). (More complicated trajectories may have a different limit or no limit.) So merely checking a few trajectories is not sufficient to prove that the limit exists. Often you need to put together an ε-δ proof to achieve this.

Answer 3

i will in basic terms teach that there are not any integer strategies. will could think of on it some extra. x^2 + y^2 + z^2 = 6 + yz + xz + xy Rearrange to quadratic variety in x : x^2 - x(y + z) + (y^2 + z^2 -yz - 6) = 0 remedy for x utilising the quadratic formulation : x = {y + z ± sqrt[(-(y + z))^2 - 4(y^2 + z^2 - yz - 6)]} / 2 This simplifies to : x = {y + z ± sqrt[3(8 - (y - z)^2)]} / 2 Now the expression below the sq. root could be extra suitable than or equivalent to 0. i.e. 8 - (y - z)^2 ? 0 i.e 8 ? (y - z)^2 i.e. (y - z)^2 ? 8 the only situations that fulfill this are y - z = 0 or ± a million or ± 2. Substituting those into sqrt[3(8 - (y - z)^2)] supplies sqrt(24) or sqrt(21) or sqrt(12), none of which yield an integer. for this reason, there are not any integer strategies.

Answer 4

as (x,y,z)-->(0,0,0), x-->y-->z, x=y=z
(xy+yz+zx)/(x^2+y^2+z^2) becomes

(x^2 + x^2 + x^2)/(3x^2)
3x^2/3x^2
1

The limit is 1

Answer 5

No. Do it yourself, idiot.