Multivariable calculus

Limit of xy/(x^2+y^2) at the Origin Does Not Exist

Q: Is the function continuous anywhere?

Yes. Everywhere except the origin the denominator x^2 + y^2 is positive, so f is a quotient of polynomials with nonvanishing denominator and is continuous there. Only the origin is a problem, and no value assigned to f(0,0) can repair it.

Q: Does approaching along a curve instead of a line change anything?

Not here. At every point of polar angle \theta the function is exactly \tfrac{1}{2}\sin 2\theta, so a curve that arrives at the origin with a definite direction gives whatever that direction's line already gives. Lines exhibit every value f can approach, from -\tfrac{1}{2} to \tfrac{1}{2}.

The limit of xy/(x^2 + y^2) as (x, y) approaches (0, 0) does not exist. Along the line y = 0 the function is 0 at every point, and along y = x it is 1/2 at every point. Two paths into the origin give two different values, so no single number can be the limit.

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

The limit does not exist

Path	Limit along it
along y = 0	0
along y = x	0.5
along y = 2x	0.4

The two-path test, done once

Two paths that disagree are all it takes to kill a limit. Substitute the path into $f$ , simplify, then let the parameter go to zero. Start with the $x$ -axis, where $y = 0$ .

f(t, 0) = \frac{t \cdot 0}{t^2 + 0^2} = 0 \quad \text{for every } t \neq 0

So the limit along that path is $0$ . Now take the diagonal $y = x$ .

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

The function is constantly $1/2$ on that path, no matter how close to the origin you get. Since $0 \neq \tfrac{1}{2}$ , the limit does not exist and you can stop.

The whole line family shows the same thing at once. Putting $y = mx$ gives a value that depends on the slope:

f(x, mx) = \frac{x \cdot mx}{x^2 + m^2x^2} = \frac{m}{1 + m^2}

$m = 0$ gives $0$ , the $x$ -axis.
$m = 1$ gives $\tfrac{1}{2}$ , the largest value any line produces.
$m = 2$ gives $\tfrac{2}{5}$ .
$m = -1$ gives $-\tfrac{1}{2}$ , the smallest.

Polar coordinates make the failure visible

Write $x = r\cos\theta$ and $y = r\sin\theta$ . The numerator and denominator are both degree two, so every power of $r$ cancels.

f = \frac{r^2\cos\theta\sin\theta}{r^2} = \cos\theta\sin\theta = \frac{1}{2}\sin 2\theta

There is no $r$ left. The value of $f$ depends only on the direction you came from, and it never settles down as $r \to 0$ : it stays pinned at $\tfrac{1}{2}\sin 2\theta$ forever. That is the signature of a path-dependent limit.

Compare this with a limit that does exist, where the polar form looks like $r \cdot (\text{something bounded})$ . There the $r$ factor crushes everything to zero regardless of $\theta$ . Here no such factor appears.

The mistake: checking only the two axes

Both axes give $0$ , so students report the limit as $0$ and move on. Agreement along a few paths is evidence, never proof. A limit exists only if every possible approach gives the same value, and there are infinitely many approaches.

Use path tests in one direction only: to disprove. When two paths agree, that is a hint to go looking for a real proof, using polar coordinates or a squeeze estimate. When two paths disagree, you have a complete argument and you are finished.

Two agreeing paths prove nothing about the limit.
Two disagreeing paths prove the limit does not exist.
A single path can never establish existence, no matter how general it looks.

Frequently asked questions

Is the function continuous anywhere?

Yes. Everywhere except the origin the denominator $x^2 + y^2$ is positive, so $f$ is a quotient of polynomials with nonvanishing denominator and is continuous there. Only the origin is a problem, and no value assigned to $f(0,0)$ can repair it.

Does approaching along a curve instead of a line change anything?

Not here. At every point of polar angle $\theta$ the function is exactly $\tfrac{1}{2}\sin 2\theta$ , so a curve that arrives at the origin with a definite direction gives whatever that direction's line already gives. Lines exhibit every value $f$ can approach, from $-\tfrac{1}{2}$ to $\tfrac{1}{2}$ .