Show that $x-\alpha f(x)\to\pm\infty$ as $x\to\pm\infty$ if $\operatorname{sgn}(\alpha)\alpha f'\le c$ for some $c>0$ Let $\alpha\in\mathbb R$, $c>0$, $f\in C^1(\mathbb R)$ with $$f'\left.\begin{cases}\le c&\text{, if }\alpha\ge0\\\ge-c&\text{, if }\alpha\le0\end{cases}\right\}\tag1$$ and $$g(x):=x-\alpha f(x)\;\;\;\text{for }x\in\mathbb R.$$

How can we show that $g(x)\to\pm\infty$ as $x\to\pm\infty$?

While this should be easy, I don't know how we need to argue. Since the first term $x$ in the definition of $g(x)$ trivially goes to $\pm\infty$ as $x\to\pm\infty$, the reasoning should be that the second term cannot grow as fast and this should somehow follow from $(1)$ ...
 A: The conclusions holds if and only if $|\alpha| c < 1$.
I'll demonstrate this for the case $\alpha > 0$, the case $\alpha < 0$ works similarly, and in the case $\alpha = 0$ we have $g(x)= x$ so that the conclusion holds trivially.
If $\alpha > 0$ and $\alpha c \ge 1$ then $f(x) = x/\alpha$ with $f'(x) = 1/\alpha  \le c$ and $g(x) = 0$ is a counterexample.
If $\alpha > 0$ and $\alpha c < 1$ then $f'(x) \le c$ implies
$$
 f(x) \le f(0) + cx \implies 
g(x) \ge -\alpha f(0) + \underbrace{(1-\alpha c)}_{> 0} \underbrace{(x)}_{> 0}
$$
for $ x > 0$, and
$$
 f(x) \ge f(0) + cx \implies 
g(x) \le -\alpha f(0) + \underbrace{(1-\alpha c)}_{> 0} \underbrace{(x)}_{< 0}
$$
for $x < 0$. It follows that
$$
 \lim_{x\to +\infty} g(x) = +\infty \, ,\,  
\lim_{x\to -\infty} g(x) = -\infty \, .
$$
A: This follows from a simple analysis: for instance, if $\alpha > 0$, you know that $f'$ is upper bounded (by a positive constant $c$), so $g(x) \geq x - \alpha c \to \infty$ as $x \to \infty$. The analysis for the other case (i.e., $\alpha < 0$) is completely similar.
A: As the comment by Martin R mentions, it doesn't hold. You need some additional criteria.
Also I'm using the condition you mention in the title which is $$\alpha f'(x) \le sgn(\alpha)c $$ which is different from what you say in $(1)$.
If we additionally have that $sgn(\alpha)c <1$, then $1 - \alpha f'(x) = 1 - sgn(\alpha)c + sgn(\alpha)c - \alpha f'(x) \ge 1 - sgn(\alpha)c >0$
Then by the Mean Value Theorem, $g(x) -g(0) = x g'(b) $ for some $b$ between $0$ and $x$.
\begin{align}
&g(x) -g(0) = x g'(b) = x(1 - \alpha f'(b))\\
\implies &g(x) = -\alpha f(0) + x(1 - \alpha f'(b)) \\
\end{align}
If $x > 0$, then
$$g(x) = -\alpha f(0) + x(1 - \alpha f'(b)) > -\alpha f(0) + x(1 - sgn(\alpha)c) $$
and we can say that $\lim_{x \to \infty} g(x)  = \infty $.
If $x <0$, we have
$$g(x) = -\alpha f(0) + x(1 - \alpha f'(b)) < -\alpha f(0) + x(1 - sgn(\alpha)c) $$
and $\lim_{x \to -\infty} g(x) = -\infty$
