Injectivity of analytic functions

Question

Suppose $f : \mathbb{R} \rightarrow \mathbb{R}^n$ is a real analytic function on $(a, \infty)$. I have two questions:

1. Suppose $||f(x)|| \rightarrow \infty$ as $x \rightarrow \infty$. I know without further conditions $f$ can be not injective (for e.g. take $f(x) = x + 2 \sin(x)$ for $n=1$). I want to know if any generic conditions are known such that $f$ is eventually injective (meaning that there exists $b > a$ s.t $f$ is injective on $(b, \infty)$.

2. Suppose $f(x)$ converges as $x \rightarrow \infty$, and $(\partial^m f / \partial x^m)(x) \rightarrow 0$ as $x \rightarrow \infty$, for all $m \geq 1$. Again without further conditions $f$ is not eventually injective (for e.g. $f(x) = e^{-x} \sin (x)$ shows it is not true generically). So again my question is whether there are some simple conditions known that makes $f$ eventually injective.

Answer 1

For $n=1$, such simple (non-tautological) conditions do not exist, because real-analytic functions can mimic any $C^1$ functions in terms of their monotonicity patterns and the limit value at $\infty-$, simultaneously. So, the real-analyticity condition does not help at all; it is far from any kind of a rigidity condition.

Indeed, for any $a\in\mathbb R$, take any $C^1$ function $g\colon[a,\infty)\to\mathbb R$ such that $\exists\ g(\infty-)\in[-\infty,\infty]$ and for some increasing to $\infty$ sequence $(x_j)$ in $[a,\infty)$ we have $(-1)^j g'(x_j)>0$ for all $j$. Extend $g$ to a $C^1$ function on $\mathbb R$ and denote the extension still by $g$.

Take now any continuous function $h\colon\mathbb R\to(0,\infty)$ such that $h(x_j)<|g'(x_j)|$ for all $j$ and $h(\infty-)=0$. Then, according to a theorem by Whitney (see e.g. the first paragraph of this paper), there is a real-analytic function $f\colon\mathbb R\to\mathbb R$ such that $|f-g|+|f'-g'|\le h$ and hence $f(\infty-)=g(\infty-)$ and $(-1)^j f'(x_j)>0$ for all $j$ -- so that the real-analytic function $f$ indeed mimics the monotonicity pattern of the $C^1$ function $g$ and has the same limit value at $\infty-$.