Multivariable calculus

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

Q: Would the limit exist if the denominator were x^2 + y^2 raised to a smaller power?

With \frac{\sin(xy)}{\sqrt{x^2+y^2}} the limit is 0, because |xy| \leq \tfrac{1}{2}(x^2+y^2) makes the quotient bounded by \tfrac{1}{2}\sqrt{x^2+y^2}. The trouble here is that numerator and denominator are the same degree.

The limit of sin(xy)/(x^2 + y^2) at the origin does not exist. Along y = 0 the function is 0, along y = x it tends to 1/2, and along y = -x it tends to -1/2. Near the origin sin(xy) behaves like xy, so this function inherits the path dependence of xy/(x^2 + y^2).

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

The limit does not exist

Path	Limit along it
along y = 0	0
along y = x	0.5
along y = -x	-0.5

Reduce the sine, then run the path test

The one-variable fact $\lim_{u \to 0} \frac{\sin u}{u} = 1$ does the heavy lifting. Write the function so that the sine appears over its own argument.

\frac{\sin(xy)}{x^2 + y^2} = \frac{\sin(xy)}{xy} \cdot \frac{xy}{x^2 + y^2} \quad (xy \neq 0)

The first factor tends to $1$ whenever $xy \to 0$ , which happens on any path into the origin. So the behaviour is decided entirely by the second factor, the classic path-dependent quotient.

Along $y = 0$ the numerator is $\sin 0 = 0$ , so $f$ is identically zero on the $x$ -axis and the limit along that path is $0$ . Along $y = x$ :

f(t, t) = \frac{\sin(t^2)}{2t^2} = \frac{1}{2}\cdot\frac{\sin(t^2)}{t^2} \to \frac{1}{2}

Along $y = -x$ the argument of the sine is $-t^2$ , and since sine is odd the same computation gives $-\tfrac{1}{2}$ . Three paths, three values, so the limit does not exist.

How to see it faster

For a fixed line $y = mx$ the argument of the sine is $mx^2$ , which is tiny, so $\sin(mx^2) \approx mx^2$ to leading order. That turns the quotient into a familiar one.

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

So $f$ approaches $\frac{m}{1+m^2}$ along the line of slope $m$ , exactly the same values as $\frac{xy}{x^2+y^2}$ . Replacing a smooth function by its leading term near the origin is a fast way to guess the answer, and then you confirm it with the exact substitution shown above.

$\sin u = u + O(u^3)$ , so the cubic correction is negligible against $u$ .
The two functions $\frac{\sin(xy)}{x^2+y^2}$ and $\frac{xy}{x^2+y^2}$ have the same path values at the origin.
Their difference does tend to zero, which is why the shortcut is safe here.

The mistake: applying sin(u)/u to the wrong denominator

The common error is to write $\frac{\sin(xy)}{x^2+y^2} \to 1$ because sine over something small is one. The identity requires the denominator to be the very argument of the sine. Here the argument is $xy$ but the denominator is $x^2 + y^2$ , and those are not the same size: on the $x$ -axis $xy$ is zero while $x^2+y^2$ is not.

The safe move is the one made above: multiply and divide by $xy$ so the identity applies to a genuine $\frac{\sin u}{u}$ , and handle the leftover factor separately. When $xy = 0$ the function is simply zero, so that case needs no special treatment.

Frequently asked questions

Would the limit exist if the denominator were x^2 + y^2 raised to a smaller power?

With $\frac{\sin(xy)}{\sqrt{x^2+y^2}}$ the limit is $0$ , because $|xy| \leq \tfrac{1}{2}(x^2+y^2)$ makes the quotient bounded by $\tfrac{1}{2}\sqrt{x^2+y^2}$ . The trouble here is that numerator and denominator are the same degree.

Is f continuous everywhere except the origin?

Yes. The sine of a polynomial is continuous everywhere and the denominator is nonzero off the origin, so $f$ is continuous on the punctured plane. Only the origin has no limit.