Multivariable calculus

Limit of xy/sqrt(x^2+y^2) at the Origin

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

Only in the power of the denominator, and that changes everything. With the square root, polar coordinates leave a surviving factor of r and the limit is 0. Without it, the quotient is \cos\theta\sin\theta with no r at all, so the value depends on direction: 0 along the axes and \frac{1}{2} along y = x, and the limit does not exist.

Q: Where does the inequality 2|xy| <= x^2 + y^2 come from?

Expand \left(|x| - |y|\right)^2 = x^2 - 2|xy| + y^2 \ge 0 and move the middle term across. It is the arithmetic mean and geometric mean inequality for the two numbers x^2 and y^2, and it is worth memorising, since it settles a large fraction of limit problems at the origin.

The limit of xy / sqrt(x^2 + y^2) as (x, y) approaches the origin is 0. In polar coordinates the function is r cos(theta) sin(theta), which is at most r/2 in size. Equivalently, 2|xy| is at most x^2 + y^2, so the quotient is bounded by half the distance to the origin.

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

The limit exists

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

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

Paths give the candidate

Start with three straight lines. Each one makes the numerator quadratic in $t$ while the denominator stays linear, so a factor of $t$ is left over.

f(t,0) = \frac{0}{|t|} = 0

f(t,t) = \frac{t^2}{\sqrt{2}\,|t|} = \frac{|t|}{\sqrt{2}} \to 0

f(t,-2t) = \frac{-2t^2}{\sqrt{5}\,|t|} = \frac{-2|t|}{\sqrt{5}} \to 0

Notice the absolute value appearing from $\sqrt{t^2} = |t|$ . Forgetting it is harmless when the limit is zero, but on a problem where the answer differs for $t > 0$ and $t < 0$ it is the difference between a limit and no limit.

Two ways to bound it

In polar coordinates the numerator is degree two in $r$ and the denominator is degree one, leaving one factor of $r$ .

f = \frac{r\cos\theta \cdot r\sin\theta}{r} = r\cos\theta\sin\theta = \frac{r}{2}\sin 2\theta

\left|f\right| \le \frac{r}{2} = \frac{\sqrt{x^2+y^2}}{2} \to 0

The same bound comes out without any change of coordinates. Start from $\left(|x| - |y|\right)^2 \ge 0$ , which rearranges to $2|xy| \le x^2 + y^2$ .

\left|f\right| = \frac{|xy|}{\sqrt{x^2+y^2}} \le \frac{x^2+y^2}{2\sqrt{x^2+y^2}} = \frac{\sqrt{x^2+y^2}}{2}

Either route squeezes $f$ between $-\frac{r}{2}$ and $\frac{r}{2}$ , both of which tend to $0$ . Use whichever you find faster; the polar version is mechanical, the algebraic version avoids trigonometry entirely.

The mistake: bounding the square root the wrong way

A tempting move is to replace $\sqrt{x^2+y^2}$ in the denominator with $|x| + |y|$ , since the two look interchangeable. They are not equal, and the inequality runs one way only: $\sqrt{x^2+y^2} \le |x| + |y|$ .

Putting the larger quantity into a denominator makes the fraction smaller, so that substitution produces a lower bound on $|f|$ . A lower bound is useless for a squeeze, which needs an upper bound tending to zero. Swapping the direction of an inequality inside a fraction is the single most common way these arguments quietly break.

If you do want to use $|x| + |y|$ , put it in the numerator instead. From $|xy| \le \frac{\left(|x|+|y|\right)^2}{4}$ you would get an upper bound, though it is messier than the one above.

Frequently asked questions

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

Only in the power of the denominator, and that changes everything. With the square root, polar coordinates leave a surviving factor of $r$ and the limit is $0$ . Without it, the quotient is $\cos\theta\sin\theta$ with no $r$ at all, so the value depends on direction: $0$ along the axes and $\frac{1}{2}$ along $y = x$ , and the limit does not exist.

Where does the inequality 2|xy| <= x^2 + y^2 come from?

Expand $\left(|x| - |y|\right)^2 = x^2 - 2|xy| + y^2 \ge 0$ and move the middle term across. It is the arithmetic mean and geometric mean inequality for the two numbers $x^2$ and $y^2$ , and it is worth memorising, since it settles a large fraction of limit problems at the origin.