Multivariable calculus

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

Q: Does the limit still exist with a different power in front?

Yes for every positive power. \left(x^2+y^2\right)^a\ln\left(x^2+y^2\right) tends to 0 whenever a > 0, however small a is, because u^a beats \ln u for any a > 0. At a = 0 the function is just \ln\left(x^2+y^2\right), which diverges to -\infty, and for a < 0 it diverges as well.

Q: Can the function be made continuous at the origin?

Yes. Setting f(0,0) = 0 matches the limit, so the extended function is continuous on the whole plane. Both partial derivatives at the origin also come out to 0, since \frac{f(h,0) - 0}{h} = 2h\ln|h| \to 0.

The limit of (x^2 + y^2) ln(x^2 + y^2) as (x, y) approaches the origin is 0. Writing u = x^2 + y^2 reduces it to the one variable limit of u ln(u) as u tends to 0 from above, which is 0: the factor shrinking to zero beats the logarithm diverging to minus infinity.

f(x,y) = \left(x^2+y^2\right)\ln\left(x^2+y^2\right)

The limit exists

\lim_{(x,y) \to (0,0)} \left(x^2+y^2\right)\ln\left(x^2+y^2\right) = 0

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

The paths creep to zero from below

Every line through the origin makes $x^2+y^2$ a constant times $t^2$ , so each path test is a shrinking factor multiplied by the logarithm of a shrinking factor.

f(t,0) = t^2\ln\left(t^2\right) = 2t^2\ln|t| \to 0

f(t,t) = 2t^2\ln\left(2t^2\right) \to 0

f(t,2t) = 5t^2\ln\left(5t^2\right) \to 0

All three values are negative near the origin, since the logarithm of a small positive number is negative, and they rise to $0$ from below. Convergence is slow: at distance $r = 0.01$ from the origin the value is still about $-0.0009$ , so a numerical table alone would leave you unsure.

Reduce to u ln(u) and see which factor wins

The function depends on the point only through $u = x^2+y^2$ , and $u \to 0^+$ under every approach, so the problem becomes a single one sided limit.

\lim_{u \to 0^+} u\ln u = \lim_{u \to 0^+} \frac{\ln u}{1/u} = \lim_{u \to 0^+} \frac{1/u}{-1/u^2} = \lim_{u \to 0^+} (-u) = 0

The middle step rewrites the product as a quotient of the form $\frac{-\infty}{+\infty}$ so that L'Hopital's rule applies, and one differentiation clears it. This is the standard result that any positive power beats a logarithm.

In polar coordinates the function is $r^2\ln\left(r^2\right) = 2r^2\ln r$ , with no $\theta$ in sight. The value near the origin depends only on distance, so it is constant on every circle centred at the origin.

The mistake: reading minus infinity off the logarithm

The logarithm really does diverge to $-\infty$ , so the second factor is unbounded, and it is easy to conclude that the product must be unbounded too. But a quantity tending to $0$ times a quantity tending to $\pm\infty$ is an indeterminate form: it can settle on any number, diverge, or fail to have a limit at all.

Compare $u\ln u \to 0$ with $\frac{\ln u}{u} \to -\infty$ as $u \to 0^+$ . Same logarithm, opposite outcomes, decided entirely by what it is paired with. Only a rate comparison settles the question, which is exactly what the L'Hopital step performs.

One more point worth being clear about: $f$ is defined at every point except the origin, since $x^2+y^2 > 0$ off the origin means the logarithm always has a positive argument. The single bad point is the one the limit is taken at, which is precisely the situation limits exist to handle.

Frequently asked questions

Does the limit still exist with a different power in front?

Yes for every positive power. $\left(x^2+y^2\right)^a\ln\left(x^2+y^2\right)$ tends to $0$ whenever $a > 0$ , however small $a$ is, because $u^a$ beats $\ln u$ for any $a > 0$ . At $a = 0$ the function is just $\ln\left(x^2+y^2\right)$ , which diverges to $-\infty$ , and for $a < 0$ it diverges as well.

Can the function be made continuous at the origin?

Yes. Setting $f(0,0) = 0$ matches the limit, so the extended function is continuous on the whole plane. Both partial derivatives at the origin also come out to $0$ , since $\frac{f(h,0) - 0}{h} = 2h\ln|h| \to 0$ .