Multivariable calculus

Limit of xyz/(x^2+y^2+z^2) at the Origin

Q: Does this limit exist in every dimension?

The same argument works for \frac{x_1 x_2 \cdots x_n}{x_1^2 + \cdots + x_n^2} whenever n \ge 3, because the numerator has degree n and the denominator degree 2. At n = 2 the quotient is \frac{xy}{x^2+y^2}, which has no limit: 0 along the axes, \frac{1}{2} along y = x.

Q: Is the bound |z|/2 better than the spherical one?

They are the same fact in different clothes. The bound \frac{|z|}{2} needs no change of coordinates and is quicker to write down. The spherical version makes the reason visible: three factors of the radius upstairs against two downstairs, leaving one factor of \rho to drive everything to zero.

The limit of xyz / (x^2 + y^2 + z^2) as (x, y, z) approaches the origin is 0. Since 2|xy| is at most x^2 + y^2, the quotient is bounded in size by |z|/2, which goes to 0 under any approach. In spherical coordinates the function is rho times a bounded trigonometric factor.

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

The limit exists

\lim_{(x,y,z) \to (0,0,0)} \frac{xyz}{x^2 + y^2 + z^2} = 0

Path	Limit along it
along the x axis	0
along the diagonal x = y = z	0
along y = x in the plane z = 0	0

Paths in space

Start with an axis, the main diagonal $x = y = z$ , and a line lying in the plane $z = 0$ .

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

f(t,t,t) = \frac{t^3}{3t^2} = \frac{t}{3} \to 0

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

In three variables the path problem gets worse, not better. Approaches can run along lines, curves, or spirals, and the set of directions is a whole sphere rather than a circle. Testing a few and generalising is even less defensible than it was in the plane.

A one line bound

Start from $\left(|x| - |y|\right)^2 \ge 0$ , which rearranges to $2|xy| \le x^2 + y^2$ . Feed that into the numerator.

\left|f\right| = \frac{|x||y||z|}{x^2+y^2+z^2} \le \frac{|z|\left(x^2+y^2\right)}{2\left(x^2+y^2+z^2\right)} \le \frac{|z|}{2}

The last step uses $x^2 + y^2 \le x^2+y^2+z^2$ , so that fraction is at most $1$ . Every approach to the origin forces $z \to 0$ , so the bound collapses and $f$ is squeezed to $0$ .

0 \le \left|f\right| \le \frac{|z|}{2} \to 0

Spherical coordinates give the same conclusion and show why it happens. With $x = \rho\sin\varphi\cos\theta$ , $y = \rho\sin\varphi\sin\theta$ , $z = \rho\cos\varphi$ , the numerator is degree three in $\rho$ and the denominator degree two.

f = \rho\sin^2\varphi\cos\varphi\cos\theta\sin\theta, \qquad \left|f\right| \le \rho

The mistake: checking the coordinate axes only

Along each coordinate axis two of the three variables are zero, so the numerator $xyz$ is identically zero and the function is flat at $0$ . Three effortless zeros feel conclusive and carry no information at all: the path killed the numerator before the denominator ever had a chance to matter.

A useful habit follows from that. When a path makes the numerator vanish identically, that path tells you nothing about the limit. Pick approaches where the numerator survives, such as $x = y = z = t$ above, and then close the argument with a bound that holds on a whole ball.

The same trap appears in two variables with functions like $\frac{xy}{x^2+y^2}$ , which is $0$ along both axes and $\frac{1}{2}$ along $y = x$ , and has no limit. Axis testing would have reported a limit of $0$ with total confidence.

Frequently asked questions

Does this limit exist in every dimension?

The same argument works for $\frac{x_1 x_2 \cdots x_n}{x_1^2 + \cdots + x_n^2}$ whenever $n \ge 3$ , because the numerator has degree $n$ and the denominator degree $2$ . At $n = 2$ the quotient is $\frac{xy}{x^2+y^2}$ , which has no limit: $0$ along the axes, $\frac{1}{2}$ along $y = x$ .

Is the bound |z|/2 better than the spherical one?

They are the same fact in different clothes. The bound $\frac{|z|}{2}$ needs no change of coordinates and is quicker to write down. The spherical version makes the reason visible: three factors of the radius upstairs against two downstairs, leaving one factor of $\rho$ to drive everything to zero.