Multivariable calculus

Limit of (x^4-y^4)/(x^2+y^2) at the Origin

Q: Does the same trick work for (x^4+y^4)/(x^2+y^2)?

Not by factoring, because x^2+y^2 does not divide x^4+y^4. Use polar instead: the quotient becomes r^2\left(\cos^4\theta + \sin^4\theta\right), and since that trigonometric factor lies between \frac{1}{2} and 1, the value is squeezed to 0.

Q: Why do the path tests give 0 so easily here?

Because after cancelling, f is a continuous function, so every path is bound to give 0. When a rational expression simplifies to something continuous, the limit is never in doubt; the real work is confirming that the simplification is exact rather than wishful.

The limit of (x^4 - y^4) / (x^2 + y^2) as (x, y) approaches the origin is 0. Because x^4 - y^4 factors as (x^2 - y^2)(x^2 + y^2), the quotient equals x^2 - y^2 at every point other than the origin. That is a polynomial, so its limit is its value there, namely 0.

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

The limit exists

\lim_{(x,y) \to (0,0)} \frac{x^4 - y^4}{x^2 + y^2} = 0

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

Factor the numerator exactly

This one is easier than it looks. The difference of two squares runs twice: $x^4 - y^4 = \left(x^2\right)^2 - \left(y^2\right)^2$ .

x^4 - y^4 = \left(x^2 - y^2\right)\left(x^2 + y^2\right)

f(x,y) = \frac{\left(x^2 - y^2\right)\left(x^2 + y^2\right)}{x^2 + y^2} = x^2 - y^2 \quad \text{for } (x,y) \ne (0,0)

The right side is a polynomial, hence continuous everywhere, so its limit at the origin is simply its value: $0 - 0 = 0$ . No squeeze, no polar coordinates, no inequality. The difficulty was an illusion created by how the fraction was written.

Check the paths against it. Along $y = 0$ you get $t^2$ , along $y = x$ you get exactly $0$ , along $y = 2x$ you get $-3t^2$ . All tend to $0$ , as the simplified form guarantees.

Confirming with polar coordinates

It is worth having a second, independent route. Substituting $x = r\cos\theta$ and $y = r\sin\theta$ gives a numerator of degree four against a denominator of degree two.

f = \frac{r^4\left(\cos^4\theta - \sin^4\theta\right)}{r^2} = r^2\left(\cos^4\theta - \sin^4\theta\right) = r^2\cos 2\theta

The last equality uses the same difference of squares: $\cos^4\theta - \sin^4\theta = \left(\cos^2\theta - \sin^2\theta\right)\left(\cos^2\theta + \sin^2\theta\right) = \cos 2\theta$ .

\left|f\right| \le r^2 \to 0

Two methods, same answer. When a result is going to be used, getting it twice by different routes is cheap insurance.

The mistake: cancelling a factor that is not there

The cancellation works here for a specific reason: $x^2+y^2$ divides $x^4-y^4$ exactly, with no remainder. Students then carry the move to $\frac{x^3-y^3}{x^2+y^2}$ or $\frac{x^2y}{x^2+y^2}$ , where the denominator is not a factor of the numerator, and cancel anyway. Nothing cancels, and the expression that comes out is a different function with a different limit.

Before cancelling, multiply back and check you recover the original numerator. Here $\left(x^2-y^2\right)\left(x^2+y^2\right) = x^4 - y^4$ , so the step is exact.

The second half of the mistake is forgetting the domain. $f$ and $x^2 - y^2$ agree at every point except the origin, where $f$ is undefined and $x^2 - y^2$ equals $0$ . That is harmless for a limit, since a limit never inspects the point itself, but the two are not the same function.

Frequently asked questions

Does the same trick work for (x^4+y^4)/(x^2+y^2)?

Not by factoring, because $x^2+y^2$ does not divide $x^4+y^4$ . Use polar instead: the quotient becomes $r^2\left(\cos^4\theta + \sin^4\theta\right)$ , and since that trigonometric factor lies between $\frac{1}{2}$ and $1$ , the value is squeezed to $0$ .

Why do the path tests give 0 so easily here?

Because after cancelling, $f$ is a continuous function, so every path is bound to give $0$ . When a rational expression simplifies to something continuous, the limit is never in doubt; the real work is confirming that the simplification is exact rather than wishful.