# 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 $(1/t^2,1/t)$?

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?

• May 23, 2020 at 6:43

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.

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 $$\{f_a(x)\}$$.

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.