If $\lim\limits_{x \to \infty} f(x) = \infty$ then a special interval exists? Sorry for the bad title, I didn't know any word to describe what I need.
Suppose $f:\mathbb R \to \mathbb R$ is $C^1$ and that $\lim\limits_{x\to \infty}f(x) = \infty$
Can we say for sure that there is some interval $[0, b]$ such that $f(x) >\max\limits_{y \in [0,b]}f(y)$ for all $x >b$?
More generally, suppose $f:\mathbb R^n \to \mathbb R^n$ a $C^1$ function such that $\lim\limits_{|x| \to \infty}|f(x)| = \infty$
Can we say for certain that there is closed ball $B(0, r)$ such that $|f(x)| > \max\limits_{y\in B(0, r)} |f(y)|$? this seems natural and trivial but I have yet to find a formal proof.
 A: The answer is no. Let us consider $f(x)=x-10\sin(\pi x)$. It can be seen easily that $\lim\limits_{x\to\infty}f(x)\ge \lim\limits_{x\to\infty}(x-10)=\infty$. Suppose any $b\ge 0$ is given. Then, there is $k\ge 0$ such that $2k\le b<2k+2$. We can see that $$f(2k)=2k,\quad f(2k+\frac52)=2k+\frac52-10\sin(\frac{\pi}2)=2k-\frac{15}2.$$ So it follows
$$
\max_{0\le y\le b}f(y)\ge f(2k)=2k> 2k-\frac{15}2=f(2k+\frac52)
$$ and $2k+\frac52>b$. This shows the desired condition does not hold for every $b\ge 0$. Note also that by defining $g(x)=|x|-10\sin(\pi|x|)+10, \ x\in\Bbb R^n$, it fails
$$
\max_{|y|\le r}|g(y)|<|g(x)|,\quad \forall |x|>r
$$ for every $r\ge 0$ for the same reason.
A: Here is a continuous counter-example (but can be smoothed easily to give a continuously differentiable example):

For all $k \in \Bbb Z_{\ge 0}$, the function $f$


*

*decreases with slope $-1$ on the interval $[5k, 5k+2]$, and 

*increases with slope $+1$ on the interval $[5k+2, 5k+5]$.


For arbitrary $b \ge 0$ there is a unique $k$ such that $5k \le b < 5k+5$. Then
$$
 \max \{ f(x) \mid 0 \le x \le b\} \ge f(5k) = k,
$$
but
$$
 f(5k+7) = k-1 \, .
$$
