Multivariable calculus

Limit of (xy+z^2)/(x^2+y^2+z^2) Does Not Exist

Q: Does this function have a limit along any plane through the origin?

Not on the planes you are likely to try. In the plane z = 0 it reduces to \frac{xy}{x^2+y^2}, which has no limit at the origin, and in the plane y = x the point (t, t, s) gives \frac{t^2 + s^2}{2t^2 + s^2}, which is \tfrac{1}{2} when s = 0 and 1 when t = 0. A plane still holds a whole family of directions, so the two-path argument runs inside it.

The limit of (xy + z^2)/(x^2 + y^2 + z^2) at the origin in three variables does not exist. Along the x-axis the function is 0, along the line y = x with z = 0 it is 1/2, and along the z-axis it is 1. Three paths give three values, so no limit can exist.

f(x,y,z) = \frac{xy + z^2}{x^2 + y^2 + z^2}

The limit does not exist

Path	Limit along it
along the x-axis	0
along y = x in the plane z = 0	0.5
along the z-axis	1

Three variables, same two-path test

Nothing about the method changes in three dimensions. Pick a curve through the origin, substitute, simplify, and take the parameter to zero. Start with the $x$ -axis, where $y = z = 0$ .

f(t, 0, 0) = \frac{0 + 0}{t^2} = 0

Now the $z$ -axis, where the numerator is $z^2$ and the denominator is also $z^2$ .

f(0, 0, t) = \frac{t^2}{t^2} = 1

That is already two different values, so the limit does not exist. A third path adds a value in between: take $y = x$ inside the plane $z = 0$ .

f(t, t, 0) = \frac{t^2}{2t^2} = \frac{1}{2}

Both numerator and denominator are homogeneous of degree two, so on every ray from the origin the function is constant. That constant depends on the direction, which is the whole problem.

Spherical coordinates give the full picture

Put $x = \rho\sin\phi\cos\theta$ , $y = \rho\sin\phi\sin\theta$ , $z = \rho\cos\phi$ . The denominator is $\rho^2$ and the numerator carries $\rho^2$ as well, so the radius cancels.

f = \sin^2\phi\cos\theta\sin\theta + \cos^2\phi

There is no $\rho$ , so the value never changes as you slide toward the origin along a fixed direction. Reading off the special directions is now easy.

$\phi = 0$ points along the $z$ -axis and gives $\cos^2\phi = 1$ .
$\phi = \pi/2$ , $\theta = 0$ points along the $x$ -axis and gives $0$ .
$\phi = \pi/2$ , $\theta = \pi/4$ gives $\tfrac{1}{2}$ .
$\phi = \pi/2$ , $\theta = -\pi/4$ gives $-\tfrac{1}{2}$ , the smallest value.

The function takes every value from $-\tfrac{1}{2}$ to $1$ at points arbitrarily close to the origin.

The mistake: testing only the three coordinate axes

In three variables the reflex is to check the $x$ -, $y$ -, and $z$ -axes and stop. Here that happens to work, because the $x$ -axis and $z$ -axis already disagree. But the reflex is dangerous in general: had the numerator been $xy$ alone, all three axes would give $0$ and you would still need the diagonal $y = x$ to find the value $\tfrac{1}{2}$ .

Approaching a point in space means approaching from a two-parameter family of directions, plus every curve. Axes are three of those directions. Treat them as a first probe, never as a proof.

When the axes do agree, switch to spherical coordinates or an inequality. If the $\rho$ -free part still depends on the angles, the limit does not exist; if a positive power of $\rho$ survives in front of a bounded factor, the limit is zero.

Frequently asked questions

Does this function have a limit along any plane through the origin?

Not on the planes you are likely to try. In the plane $z = 0$ it reduces to $\frac{xy}{x^2+y^2}$ , which has no limit at the origin, and in the plane $y = x$ the point $(t, t, s)$ gives $\frac{t^2 + s^2}{2t^2 + s^2}$ , which is $\tfrac{1}{2}$ when $s = 0$ and $1$ when $t = 0$ . A plane still holds a whole family of directions, so the two-path argument runs inside it.

How would the answer change if the numerator were xyz?

Then the numerator is degree three against a degree-two denominator, and $|xyz| \leq \rho^3$ gives $|f| \leq \rho$ , so the limit would be $0$ . Mismatched degrees make the limit exist; matched degrees are what let direction matter.