Multivariable calculus

Limit of x^2y^2/(x^4+y^4) at (0,0) Does Not Exist

Q: How does this differ from x^2y^2/(x^2+y^2)?

That one has limit 0. Its numerator is degree four and denominator degree two, leaving a factor of r^2 after cancelling in polar form, and r^2 \to 0 regardless of direction. Matching degrees is what causes path dependence.

Q: What is the largest value f takes near the origin?

Exactly \tfrac{1}{2}, attained on the lines y = x and y = -x. Since \frac{m^2}{1+m^4} \leq \tfrac{1}{2} for all m, with equality at m = \pm 1, the function ranges over [0, \tfrac{1}{2}] arbitrarily close to the origin.

The limit of x^2 y^2/(x^4 + y^4) at the origin does not exist. Along y = 0 the function is 0 at every point and along y = x it is 1/2 at every point. All terms are even and the function is nonnegative, yet the value still depends on the direction of approach.

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

The limit does not exist

Path	Limit along it
along y = 0	0
along y = x	0.5
along y = 2x	0.23529411764705882

Equal degrees mean the lines decide

Numerator and denominator are both degree four, so on a line every power of $x$ cancels and the value is a pure function of the slope.

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

Take two slopes and compare. On the $x$ -axis, $m = 0$ gives $0$ . On the diagonal, $m = 1$ gives $\tfrac{1}{2}$ .

f(t, 0) = \frac{0}{t^4} = 0, \qquad f(t, t) = \frac{t^4}{2t^4} = \frac{1}{2}

Two paths disagree, so the limit does not exist. The line $y = 2x$ gives a third value, $\frac{4}{17}$ , which is about $0.235$ . Every value in $[0, \tfrac{1}{2}]$ appears on some line, since $\frac{m^2}{1+m^4}$ is maximised at $m = \pm 1$ .

Polar coordinates confirm it

Substituting $x = r\cos\theta$ , $y = r\sin\theta$ cancels $r^4$ from both levels of the fraction.

f = \frac{\cos^2\theta \sin^2\theta}{\cos^4\theta + \sin^4\theta}

No $r$ survives, so the function is constant on every ray. At $\theta = 0$ it is $0$ and at $\theta = \pi/4$ it is $\tfrac{1}{2}$ , matching the two path computations. The surface is ruled by straight lines through the origin, each one held at its own height, so no single height can serve as the limit.

The mistake: assuming a nonnegative function must go to zero

Every term here is an even power, so $f \geq 0$ and the numerator $x^2y^2$ looks like it vanishes faster than the denominator. That intuition compares the numerator to a single variable's fourth power, but the denominator is a sum, and on the diagonal the two are the same size.

A second version of the same error is to say $x^2y^2 \leq x^4 + y^4$ implies $f$ is small. The inequality is true, and it only gives $0 \leq f \leq 1$ . Boundedness is not convergence.

Compare total degrees: numerator degree four, denominator degree four, so no shrinking factor is left over.
A limit of zero needs the numerator degree to exceed the denominator degree, not merely to look small.
Nonnegative and bounded still allows the value to depend on direction.

Frequently asked questions

How does this differ from x^2y^2/(x^2+y^2)?

That one has limit $0$ . Its numerator is degree four and denominator degree two, leaving a factor of $r^2$ after cancelling in polar form, and $r^2 \to 0$ regardless of direction. Matching degrees is what causes path dependence.

What is the largest value f takes near the origin?

Exactly $\tfrac{1}{2}$ , attained on the lines $y = x$ and $y = -x$ . Since $\frac{m^2}{1+m^4} \leq \tfrac{1}{2}$ for all $m$ , with equality at $m = \pm 1$ , the function ranges over $[0, \tfrac{1}{2}]$ arbitrarily close to the origin.