Multivariable calculus

Limit of (x^2+y^2)sin(1/(x^2+y^2)) at the Origin

Q: Is f continuous at the origin once you define f(0,0) = 0?

Yes. The limit is 0, so the extended function is continuous there, and it is continuous elsewhere already. It is a standard example of a function that is continuous at a point while oscillating infinitely often in every neighbourhood of it.

The limit of (x^2 + y^2) sin(1 / (x^2 + y^2)) as (x, y) approaches the origin is 0. The sine factor oscillates between -1 and 1 forever and has no limit, but it is bounded, so the whole function is trapped between minus and plus (x^2 + y^2) and the squeeze theorem forces it to 0.

f(x,y) = \left(x^2+y^2\right)\sin\left(\frac{1}{x^2+y^2}\right)

The limit exists

\lim_{(x,y) \to (0,0)} \left(x^2+y^2\right)\sin\left(\frac{1}{x^2+y^2}\right) = 0

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

The oscillating factor never settles

Look at the second factor on its own. As the point nears the origin, $\frac{1}{x^2+y^2}$ runs to $+\infty$ , so $\sin\left(\frac{1}{x^2+y^2}\right)$ swings between $-1$ and $1$ infinitely many times inside any disc around the origin, however small that disc is. That factor has no limit and never will.

The paths still behave, because the first factor is doing the work.

f(t,0) = t^2\sin\left(\frac{1}{t^2}\right), \qquad \left|f(t,0)\right| \le t^2 \to 0

f(t,t) = 2t^2\sin\left(\frac{1}{2t^2}\right), \qquad \left|f(t,t)\right| \le 2t^2 \to 0

Along each path the value keeps changing sign all the way in, yet the amplitude of the wobble shrinks to nothing. That is what convergence looks like for an oscillating function.

The squeeze theorem in its cleanest form

The sine of any real number lies in $[-1,1]$ . That single fact is the entire proof.

-\left(x^2+y^2\right) \le \left(x^2+y^2\right)\sin\left(\frac{1}{x^2+y^2}\right) \le x^2+y^2

Both outer expressions tend to $0$ as $(x,y) \to (0,0)$ , so the middle one has nowhere else to go. In polar form the bound reads $\left|f\right| \le r^2$ , with no $\theta$ appearing, so it is uniform in direction.

This is the two variable version of the familiar $x^2\sin\frac{1}{x}$ , and it works for the same reason: a factor shrinking to zero multiplied by a factor that merely stays bounded.

The mistake: no limit inside means no limit outside

The usual argument against this limit is that $\sin\left(\frac{1}{x^2+y^2}\right)$ has no limit, so a product involving it cannot have one either. The product rule for limits does require both factors to converge, so that rule is unavailable here. But a rule failing to apply is not evidence that its conclusion is false.

What decides the outcome is that the troublesome factor is not merely limitless: it is bounded. Bounded times shrinking is shrinking, every time.

Change the pairing and the outcome changes. Replace $x^2+y^2$ in front by the constant $1$ and the product oscillates forever with no limit. Replace the sine by $\frac{1}{x^2+y^2}$ , which is unbounded, and the product is the constant $1$ with limit $1$ . The behaviour is a property of the pair, not of either factor alone.

Frequently asked questions

Is f continuous at the origin once you define f(0,0) = 0?

Yes. The limit is $0$ , so the extended function is continuous there, and it is continuous elsewhere already. It is a standard example of a function that is continuous at a point while oscillating infinitely often in every neighbourhood of it.

Does the squeeze still work if sin is replaced by cos?

Yes, and for anything bounded in that slot. The argument only uses $\left|\sin\left(\frac{1}{x^2+y^2}\right)\right| \le 1$ , so $\cos$ , or any function whose values stay inside a fixed interval, gives exactly the same conclusion.