Limit of x^2y/(x^4+y^2): Lines Agree, Limit Still Fails

Match the degrees, then choose the path

Start with lines, because they are what everyone tries first. Put $y = mx$ into $f$ and simplify.

For any fixed $m \neq 0$ this goes to $\frac{0}{m^2} = 0$ as $x \to 0$. The $x$-axis, where $m = 0$, has numerator zero throughout, and so does the vertical line $x = 0$. So every line agrees on $0$, and a student who stops here concludes the limit is $0$.

Look at the denominator to see why lines are hopeless. The two terms $x^4$ and $y^2$ balance when $y$ is comparable to $x^2$, not to $x$. On a line, $y^2 = m^2x^2$ swamps $x^4$ near the origin, so the fraction is really $\frac{mx^3}{m^2x^2}$, which vanishes. Choose the path that makes both terms the same size instead.

The function is exactly $\tfrac{1}{2}$ at every point of the parabola $y = x^2$, arbitrarily close to the origin. Since lines give $0$ and this parabola gives $\tfrac{1}{2}$, the limit does not exist.

The rule of thumb for picking the killer path

Read the denominator and ask which curve makes its two terms comparable. Here $x^4$ and $y^2$ are equal in size along $y = \pm x^2$, so those are the candidate paths. The family $y = kx^2$ shows the whole spread of values.

• $k = 1$ gives $\tfrac{1}{2}$, the largest value.
• $k = -1$ gives $-\tfrac{1}{2}$, the smallest.
• $k \to 0$ recovers the value $0$ that the lines produced.
• Any two different values of $k$ already disprove the limit.

Polar coordinates are less helpful here than usual. The function becomes $\frac{r\cos^2\theta\sin\theta}{r^2\cos^4\theta + \sin^2\theta}$, and for fixed $\theta$ with $\sin\theta \neq 0$ this tends to $0$. Holding $\theta$ fixed is just the line test wearing different clothes, so it misses the parabola too.

The mistake: believing that all lines is the same as all paths

This function exists in every textbook for one reason: it breaks the belief that checking lines is enough. Approaching along all straight lines gives $0$, and yet the limit does not exist. There is no finite list of paths whose agreement proves existence.

The practical habit: after lines agree, do not write the limit down. Either find a curve that balances the denominator, or prove the limit with an inequality. For this function the search succeeds, because $|f|$ reaches $\tfrac{1}{2}$ on points as close to the origin as you like.

The same trap appears with $\frac{xy^2}{x^2 + y^4}$, where the killer path is the sideways parabola $x = y^2$, and with $\frac{x^3y}{x^6 + y^2}$, where it is the cubic $y = x^3$. In each case the winning path is the one that makes the denominator's two terms the same order.