Multivariable calculus

Limit of xy^2/(x^2+y^4) at the Origin Does Not Exist

Q: How close to the origin does f still equal 1/2?

At every point of the curve x = y^2 with y \neq 0. Taking y = 10^{-6} gives the point (10^{-12}, 10^{-6}), which is a millionth of a unit from the origin, and f there is exactly \tfrac{1}{2}.

Q: Would polar coordinates find this?

Not if you hold the angle fixed. In polar form f = \frac{r\cos\theta\sin^2\theta}{\cos^2\theta + r^2\sin^4\theta}, which tends to 0 for each fixed \theta with \cos\theta \neq 0. Fixing \theta is the line test again, so it misses the parabola.

The limit of x y^2/(x^2 + y^4) at the origin does not exist. Every straight line through the origin gives 0, but the sideways parabola x = y^2 gives 1/2 at every point. Two disagreeing paths are all you need, so the limit fails.

f(x,y) = \frac{x y^2}{x^2 + y^4}

The limit does not exist

Path	Limit along it
along x = 0	0
along y = x	0
along the parabola x = y^2	0.5

Lines first, then the curve that balances the denominator

Substitute $y = mx$ and simplify. Cancel the largest common power of $x$ before deciding anything.

f(x, mx) = \frac{x \cdot m^2x^2}{x^2 + m^4x^4} = \frac{m^2x^3}{x^2(1 + m^4x^2)} = \frac{m^2 x}{1 + m^4x^2}

As $x \to 0$ the numerator goes to $0$ and the denominator goes to $1$ , so every line gives $0$ . The vertical line $x = 0$ gives $0$ too, since the numerator is identically zero there.

Now read the denominator: $x^2$ and $y^4$ are the same size when $x$ is comparable to $y^2$ . That points at the sideways parabola $x = y^2$ , which is the path lines cannot imitate.

f(t^2, t) = \frac{t^2 \cdot t^2}{(t^2)^2 + t^4} = \frac{t^4}{2t^4} = \frac{1}{2}

So $f$ equals $\tfrac{1}{2}$ at every point of $x = y^2$ except the origin itself. Lines give $0$ , this parabola gives $\tfrac{1}{2}$ , and the limit does not exist.

The mistake: reaching for y = x^2 out of habit

Students who have seen $\frac{x^2y}{x^4+y^2}$ try $y = x^2$ here by reflex. Watch what that gives.

f(t, t^2) = \frac{t \cdot t^4}{t^2 + t^8} = \frac{t^5}{t^2(1 + t^6)} = \frac{t^3}{1 + t^6} \to 0

Zero again, and the student concludes the limit is $0$ . The parabola has to open along the correct axis. In this function the fourth power sits on $y$ , so the curve must have $x$ growing like $y^2$ , not $y$ growing like $x^2$ .

Locate the variable carrying the higher power in the denominator: here it is $y$ , with $y^4$ .
Set the other variable equal to that variable squared: $x = y^2$ .
Substitute and check that both denominator terms become the same power of $t$ .

The whole family $x = cy^2$ with $c \neq 0$ gives $\frac{c}{c^2+1}$ , so every nonzero $c$ produces a nonzero value that disagrees with the lines.

What the surface actually looks like

Along each parabola $x = cy^2$ the function is a constant, so the surface is built from ridges and troughs that all funnel into the origin at different heights. The highest ridge sits at $\tfrac{1}{2}$ on $x = y^2$ and the deepest trough at $-\tfrac{1}{2}$ on $x = -y^2$ .

Because those ridges pass arbitrarily close to the origin, any disc around the origin contains points where $f$ is near $\tfrac{1}{2}$ and points where $f$ is near $-\tfrac{1}{2}$ . That is the precise reason no epsilon-delta argument can work.

Frequently asked questions

How close to the origin does f still equal 1/2?

At every point of the curve $x = y^2$ with $y \neq 0$ . Taking $y = 10^{-6}$ gives the point $(10^{-12}, 10^{-6})$ , which is a millionth of a unit from the origin, and $f$ there is exactly $\tfrac{1}{2}$ .

Would polar coordinates find this?

Not if you hold the angle fixed. In polar form $f = \frac{r\cos\theta\sin^2\theta}{\cos^2\theta + r^2\sin^4\theta}$ , which tends to $0$ for each fixed $\theta$ with $\cos\theta \neq 0$ . Fixing $\theta$ is the line test again, so it misses the parabola.