Unbounded cones where $|y|e^{-|y|}(1+x^2y)$ goes to $0$ I am given the function $f : \mathbb R^2 \to \mathbb R$ such that
$$(x,y)\mapsto f(x,y) = |y|e^{-|y|}(1+x^2y), $$
and I am asked what condition I should impose on a closed, unbounded cone $C$ with vertex in the origin such that
$$\lim_{|(x,y)|\to\infty,\ (x,y) \in C} f(x,y) = 0. $$
To answer this problem I first restricted myself to rays $(\alpha t,\beta t)$ with $\alpha = \cos\theta$, $\beta = \sin\theta$ and $\theta \in [0,2\pi)$, $t \geq 0$. By tackling the two cases $\beta > 0$, $\beta < 0$ one sees that, thanks to the exponential, $|f|$ decays to $0$ along all rays, irrespective of $\alpha$. When $\beta = 0$ the function is trivially zero, so the same holds.
However, we aren't allowed to conclude that all cones are valid, since in general the fact that a two-variable function behaves some way along all rays does not imply that there isn't a pesky sequence leading to a different result. And indeed, when the ray is very close to the $x$-axis on either side, the restriction of $|f|$ to said ray has a maximum (due to Weierstrass's theorem: $f$ is continuous and goes to $0$ as $t \to \infty$) that increases without bound as $\theta \to 0^+, \pi, 2\pi^-$, which means that if I select the pesky sequence
$$(x_n,y_n) = (t_n\cos(1/n), t_n\sin(1/n)),$$
with $t_n$ being the point where $|f|$ attains said maximum, then $|f(x_n,y_n)| \to \infty$. So this suggest that we should prescribe that $C$ to avoid the $x$-axis entirely.
I want to justify the fact that there is an increasing maximum more concretely than just with intuition. I have expanded to first order in $\theta$, for example in the case $\theta \to 0^+$: for fixed $t \geq 0$,
$$\begin{split}
f(t\cos\theta,t\sin\theta) &= t(\sin\theta)e^{-t\sin\theta} (1 + t^3\cos^2 \theta \sin\theta) \\
&\approx \theta t (1 - \theta t) (1+\theta t^3) \approx \theta t,
\end{split}$$
which I reckon means that as $\theta$ approaches $0^+$, the function along the ray is more and more similar to a function that explodes linearly with $t$. However, I'm not sure about this argument and I don't see how it should help me.
Trying to explicitly maximize $|f|$ for a fixed ray is out of the question, seeing as it requires me to solve a quartic equation in $t$. 
After some tinkering I have noticed that restricting $f$ to the double hyperbola $(x, \pm 1/x)$ leads to a non-zero limit:
$$\left| f\left(x,\pm \frac 1 x\right) \right| = \left|\frac 1 x e^{-1/x} (1 \pm x)\right| \approx \left|\frac{x\pm 1}{x} \right| \xrightarrow{x\to\pm\infty} 1,  $$
which I guess goes to prove my claim (using $x_n = n$, $y_n = \pm 1/n$).
Questions:


*

*Is studying the restriction of $|f|$ to rays useful at all?

*What's an efficient way to prove that the max of $|f|$ on a ray increases without bound as the ray approaches the $x$-axis?

*Is it possible to arrive at the double hyperbola by studying the rays?

*How do I know problems arise only around the $x$-axis?

 A: Let $E$ be a closed subset of the unit circle. Define the cone $C_E=\{tz:t\ge 0, z\in E\}.$ Every cone under discussion equals $C_E$ for some $E.$
Claim: $C_E$ has the desired property iff neither of the points $(-1,0),(1,0)$ is in $E.$
Proof of claim: Note that because $E$ is closed, $E$ has the second property iff $E\subset \{e^{i\theta}: |\sin \theta |>c \}$ for some $c>0.$
Suppose $C_E$ has the first property. Assume, to reach a contradiction, that $E$ fails the second property. Then no positive $c$ works. Thus there exist $e^{i\theta_n} \in E$ such that either $\theta_n \to 0$ or $\theta_n \to \pi.$ Assume it's the former.
WLOG, $\theta_n \to 0^+.$ Then the points $(\sin \theta_n)^{-1}e^{i\theta_n}\in E.$ Note that $|(\sin \theta_n)^{-1}e^{i\theta_n}|\to \infty.$ Our function $f$ maps these points to 
$$1\cdot e^{-1}\cdot (1+ (\sin \theta_n)^{-2}\cos^2 \theta_n ).$$
This sequence $\to \infty$ as $n\to \infty.$ This contradicts our assumption on $E.$
I'll leave the proof of the converse to you for now.
