Multivariable calculus

Limit of x^2y/(x^2+y^4) at the Origin

The limit of x^2 y / (x^2 + y^4) as (x, y) approaches the origin is 0. Write the function as y times x^2/(x^2 + y^4). The second factor always lies between 0 and 1, so the size of the function is at most |y|, and any approach to the origin forces that to 0.

f(x,y) = \frac{x^2 y}{x^2 + y^4}

The limit exists

\lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^2 + y^4} = 0

Path	Limit along it
along y = 0	0
along y = x	0
along the parabola x = y^2	0

Paths, including the curve that usually breaks these

Start with the $x$ axis and the diagonal, then substitute the parabola $x = y^2$ .

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

f(t,t) = \frac{t^3}{t^2+t^4} = \frac{t}{1+t^2} \to 0

f\left(t^2, t\right) = \frac{t^4 \cdot t}{t^4 + t^4} = \frac{t}{2} \to 0

The third path is the parabola $x = y^2$ , chosen deliberately because it makes the two denominator terms the same size. On problems of this shape that is the approach most likely to expose a failure, and here it still returns $0$ . That raises confidence a lot, and still proves nothing.

The bounded factor argument

Split the function so that one piece obviously tends to zero and the other is obviously bounded.

f = y \cdot \frac{x^2}{x^2 + y^4}

The second factor is a nonnegative number divided by something at least as large, so it lands in $[0,1]$ at every point other than the origin.

0 \le \frac{x^2}{x^2+y^4} \le 1 \quad \Longrightarrow \quad \left|f\right| \le |y|

Any approach to the origin drives $|y| \to 0$ , so the squeeze gives a limit of $0$ .

Polar coordinates are a poor tool here. They give $\frac{r^3\cos^2\theta\sin\theta}{r^2\cos^2\theta + r^4\sin^4\theta}$ , where the two denominator terms carry different powers of $r$ , so no single factor of $r$ comes out cleanly. When the denominator is not homogeneous, reach for a bound rather than for polar.

The mistake: confusing this with the standard counterexample

Move the fourth power to the other variable and the answer flips completely.

g(x,y) = \frac{x^2 y}{x^4 + y^2}

For $g$ , along the parabola $y = x^2$ the value is $\frac{x^4}{2x^4} = \frac{1}{2}$ , while along $y = 0$ it is $0$ . Two paths, two values, so $g$ has no limit at the origin. The two functions look nearly identical and behave in opposite ways.

The structural difference is which terms match. In $f$ the denominator contains $x^2$ , exactly the factor sitting in the numerator, so that pair divides out of the size estimate and only the lone $y$ is left over. In $g$ the denominator's terms are $x^4$ and $y^2$ , neither of which matches $x^2y$ , and along $y = x^2$ the numerator and denominator are both of size $x^4$ , which lets a nonzero value survive all the way in.

Frequently asked questions

How do I know which curved path to try?

Match the degrees. Setting the two denominator terms equal in size, $x^2 = y^4$ , suggests $x = y^2$ . That is the approach along which the denominator is as small as it can be relative to its parts, so if any path is going to produce a nonzero value, it is that one.

If every path I try gives 0, is the limit 0?

Only if you check every path, which cannot be done one at a time. That is why the bound $\left|f\right| \le |y|$ matters: it holds at every point near the origin simultaneously, so it settles all approaches in a single line.