Multivariable calculus

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

Q: Why does the bound |f| <= y^2 succeed where the path tests could not?

Because it holds at every point of a punctured disc around the origin, all at once, rather than along one curve. Once 0 \le f \le y^2 everywhere nearby, any approach whatsoever forces y \to 0 and therefore f \to 0.

Q: Could I have bounded by x^2 instead?

Yes. By symmetry, f = x^2 \cdot \frac{y^2}{x^2+y^2} \le x^2, and x^2 \to 0 as well. Either bound closes the argument on its own; there is no need to write both.

The limit of x^2 y^2 / (x^2 + y^2) as (x, y) approaches the origin is 0. Split the function as y^2 times x^2/(x^2 + y^2). The second factor always lies between 0 and 1, so the whole thing is at most y^2, which is squeezed to 0 under any approach.

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

The limit exists

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

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

The paths, and what they are worth

Start with three lines through the origin. The numerator is degree four against a denominator of degree two, so each substitution leaves a $t^2$ behind.

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

f(t,t) = \frac{t^4}{2t^2} = \frac{t^2}{2} \to 0

f(t,2t) = \frac{4t^4}{5t^2} = \frac{4t^2}{5} \to 0

These agree on $0$ , so $0$ is the only candidate. They do not establish it, because there is no finite list of curves whose behaviour forces the behaviour of every other curve.

Two proofs, one with polar and one without

In polar coordinates the numerator is degree four in $r$ against a denominator of degree two, so two factors of $r$ survive.

f = \frac{r^4\cos^2\theta\sin^2\theta}{r^2} = r^2\cos^2\theta\sin^2\theta, \qquad 0 \le f \le r^2

The second proof never leaves rectangular coordinates. Peel off the piece you can bound.

f = y^2 \cdot \frac{x^2}{x^2+y^2}, \qquad 0 \le \frac{x^2}{x^2+y^2} \le 1

0 \le f \le y^2 \to 0

The fraction is between $0$ and $1$ because $x^2$ is nonnegative and no larger than $x^2 + y^2$ . Since $f \ge 0$ here, the lower half of the squeeze comes free and you only need the upper bound.

The mistake: splitting the quotient into two limits

The quotient rule for limits has a hypothesis that gets skipped: both limits must exist and the denominator's limit must be nonzero. Here the denominator tends to $0$ , so writing $\frac{\lim\left(x^2y^2\right)}{\lim\left(x^2+y^2\right)} = \frac{0}{0}$ is not an application of any theorem.

And $\frac{0}{0}$ is not a verdict. It does not mean the limit is $0$ , and it does not mean the limit fails to exist. It means the rule you tried does not apply and the rates matter. Here the numerator vanishes to fourth order and the denominator to second, so the numerator wins and the answer is $0$ . In $\frac{x^2}{x^2+y^2}$ both vanish to second order, and that limit genuinely does not exist.

Frequently asked questions

Why does the bound |f| <= y^2 succeed where the path tests could not?

Because it holds at every point of a punctured disc around the origin, all at once, rather than along one curve. Once $0 \le f \le y^2$ everywhere nearby, any approach whatsoever forces $y \to 0$ and therefore $f \to 0$ .

Could I have bounded by x^2 instead?

Yes. By symmetry, $f = x^2 \cdot \frac{y^2}{x^2+y^2} \le x^2$ , and $x^2 \to 0$ as well. Either bound closes the argument on its own; there is no need to write both.