Why does $f(x,y)= \frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?

Question

I have a stupid question about continuity in higher dimensions.

There are maps, for example, $f(x,y)=\frac{xy^2}{x^2+y^4}$, when $(x,y)\neq (0,0)$ and $f(x,y)=(0,0)$ when $(x,y)=(0,0)$, when we approach $(0,0)$ along every straight line, the limit of the function is $0$, but when along a curve, for example $(\frac{1}{t^2},\frac{1}{t})$, the limit of $f$ is not $0$.

But, it feels like all the straight lines can cover a neighbourhood of $(0,0)$, so every point on a curve is also on a different straight line. Why is it that when the same points are arranged in a different way the limit changes?

Answer 1

The reason is that the function is approaching zero along each straight line at a different speed, depending on the line. So, for $f$ to be, say, less than $1/10$ along one of the lines, you need to be within a distance $1$ from the origin, whereas for a different line you need to be within a distance of $1/2$ units, etc. It is then perfectly possible that if you approach $(0,0)$ along a curve that is transversal to all those lines, at the points of intersections with the lines your function is all the time equal to $1/10$, which makes the limit along the curve equal $1/10$. This does not contradict the fact that along all the lines the limit is zero.

Answer 2

This has something to do with the uniform continuity around 0. Given a fixed $\epsilon > 0$, if there exist an universal constant $\delta_u$ such that

$$|(x,y)|<\delta_u \implies |f(x,y)|<\epsilon$$

then your intuition is right: along any curve go to 0, $f$ will have limit 0.

However, for this example the origin is not uniformly continuous. Along lines $y=ax$, $a \in\mathbb{R}$ we can define a class of fucntions $f_a(r) = \frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $\delta_u$ exist, then for every $a$ we must have

$$ |x|<\delta_u \implies |f_a(x)|<\epsilon $$

However, let $a=\frac{1}{\sqrt{\delta_u}}$ we have $f_a(\delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to $\{f_a(x)\}$.

Answer 3

For the same reason that traveling from New York to Florida takes a few hours if you fly straight south, but takes a lot more if you choose to go first to Los Angeles.