Multivariable calculus

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

Q: Does the answer change along y = mx for different slopes m?

No. Along y = mx the function is \frac{mt^3}{\left(1+m^2\right)t^2} = \frac{mt}{1+m^2}, which tends to 0 for every m. All straight lines give 0, which is consistent with the limit being 0 but still proves nothing by itself, since curved approaches are not covered.

Q: Is the function continuous at the origin?

Not as written, since f is undefined at (0,0). But the limit exists and equals 0, so defining f(0,0) = 0 patches the hole and produces a function continuous on the whole plane.

The limit of x^2 y / (x^2 + y^2) as (x, y) approaches the origin is 0. Paths into the origin all give 0, but agreement between paths is not a proof. In polar coordinates the function becomes r cos^2(theta) sin(theta), whose size is at most r, so it is squeezed to 0 as r goes to 0.

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

The limit exists

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

Path	Limit along it
along y = 0	0
along y = x	0
along the parabola y = x^2	0

Three paths, all giving 0

Testing a path means replacing $x$ and $y$ by functions of a single parameter $t$ and letting $t \to 0$ . Along the $x$ axis the numerator is dead on arrival, so the function is identically zero.

\text{along } y = 0: \quad f(t,0) = \frac{t^2 \cdot 0}{t^2} = 0

Along the diagonal the function is not identically zero, but it still collapses.

\text{along } y = x: \quad f(t,t) = \frac{t^3}{2t^2} = \frac{t}{2} \to 0

A curve does no better than a line here, not even one that comes in tangent to the $x$ axis.

\text{along } y = x^2: \quad f\left(t, t^2\right) = \frac{t^4}{t^2 + t^4} = \frac{t^2}{1 + t^2} \to 0

Three agreements are evidence, not proof. Infinitely many curves reach the origin, and the limit exists only if every single one of them gives the same value. What the paths have bought you is a candidate: the number to try to prove.

Polar coordinates finish the job

Set $x = r\cos\theta$ and $y = r\sin\theta$ . Then $(x,y) \to (0,0)$ is exactly $r \to 0^+$ with $\theta$ left completely free, so one calculation covers every possible approach at once.

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

Now bound the trigonometric part. Since $\left|\cos^2\theta\sin\theta\right| \le 1$ for every angle, you get a bound with no $\theta$ left in it.

-r \le f \le r

Both ends go to $0$ as $r \to 0$ , and they do so at the same rate regardless of direction. That uniformity in $\theta$ is the whole point: it is what upgrades a direction-by-direction check into a statement about a full disc. The limit is $0$ .

The mistake: stopping once the paths agree

The most common error is to compute three paths, see $0$ three times, and write down the answer as proved. Paths can only ever disprove. If two disagree you are finished and the limit does not exist. If they agree you have learned only which value to aim at.

The reason is that a limit in two variables is a claim about every point in a small disc, not about a list of curves. Checking curves is like checking a handful of entries in an infinite table.

Polar coordinates are not automatically a proof either. The step that matters is that the bound $r$ carries no $\theta$ . Had the $\theta$ dependence survived without a factor of $r$ in front, as happens for $\frac{xy}{x^2+y^2} = \cos\theta\sin\theta$ , the value would depend on direction and no limit would exist.

Frequently asked questions

Does the answer change along y = mx for different slopes m?

No. Along $y = mx$ the function is $\frac{mt^3}{\left(1+m^2\right)t^2} = \frac{mt}{1+m^2}$ , which tends to $0$ for every $m$ . All straight lines give $0$ , which is consistent with the limit being $0$ but still proves nothing by itself, since curved approaches are not covered.

Is the function continuous at the origin?

Not as written, since $f$ is undefined at $(0,0)$ . But the limit exists and equals $0$ , so defining $f(0,0) = 0$ patches the hole and produces a function continuous on the whole plane.