Let $g:U\subset \mathbb R^2 \to\mathbb R$ be a function defined in a neighborhood $U$ of $(0,0)$. I'd like to understand the relationship between the following three limits.
$\displaystyle \lim_{v\to 0} \frac{g(v)}{\|v\|}=0$
$\displaystyle \lim_{v\to 0} \frac{g(tv_0)}{t}=0$, for all $v_0\in\mathbb R^2$.
$\displaystyle\lim_{t\to 0} \frac{g(\gamma(t))}{t}=0$, for all differentiable path $\gamma:(-\varepsilon,\varepsilon)\to U$ such that $\gamma(0)=(0,0)$.
Remarks:
- Limits having the form 1 appear in the definition of differentiability.
- It is clear that 1 implies 2 (what is used, for example, to prove existence of the directional derivatives of differentiable functions).
- In general, 2 does not imply 1 (what can be seen, for example, by defining $g(0,0)=0$ and $g(x,y)=\frac{x^3y}{x^6+y^2}$ otherwise).
- Of course, 2 is a much weaker version of 3 (with $\gamma$ restrict to linear paths of the form $\gamma(t)=tv_0$).
Question: Does 3 imply 1?
I couldn't prove it (for example, I'm not able to connect $|v|$ with $t$ in a $\delta$-$\epsilon$ proof; I also cannot see how could we find a $\delta$ that works for all $v$ inside a disk since each $\delta=\delta_\gamma$ works only for $v$ in the image of $\gamma$).
Also, I couldn't construct a counterexample. In fact, in every case that I have seen in which 1 fails, there are differentiable paths that do not satisfy the limit in 3.