If $\lim_{x \to \infty} f''(x) = k$, with $k > 0$, then $\lim_{x \to \infty} f(x) = \infty$ I'm having some trouble proving the following:

Let $f: \mathbb R \to \mathbb R$ be a function such that $\lim_{x \to \infty} f''(x) = k$, with $k &gt; 0$. Prove that $\lim_{x \to \infty} f(x) = \infty$.

I did the following:
If $\lim_{x \to \infty} f''(x) = k &gt; 0$, then this means that: $$\exists M\in \mathbb R: x &gt; M \implies f''(x)&gt;0$$
This means that the function $f'|_{(M,\infty)}: (M,\infty) \to \mathbb R$ with $f'|_{(M,\infty)}(x) = f'(x)$ is incresing, because it's derivative is allways $&gt;0$.
Then my plan was to do the same and then conclude that $f$ was increasing that the limit is $+\infty$. But then I remembered the function can always be increasing but no go to $\infty$ as $x \to \infty$, like, for example:  $-1/x$ for $x &gt; 0$.
How can I prove this?
$f''(x) &gt; 0$ for large $x$ is not sufficient. But you have
$$
 f''(x) \ge \frac 12 k &gt; 0
$$
for all sufficiently large $x \ge x_0$. It follows that
$$
 f'(x) \ge f'(x_0) + \frac 12 k (x-x_0)
$$
and then
$$
 f(x) \ge f(x_0) + f'(x_0) (x-x_0) + \frac 14 k (x-x_0)^2
$$
for all $x \ge x_0$. The right-hand side (the equation of an upward parabola) tends to $+\infty$ for $x \to +\infty$, which gives the desired conclusion.
We know that $f'(x)$ will be strictly increasing after sometime. If $f'(x)$ tends to a fixed value as $x$ tends to $\infty$ then clearly $\lim\limits_{x \to \infty} f''(x)=0$.  As $\lim\limits_{x \to \infty} f''(x) &gt; 0$ therefore $f'(x)$ does not tends to a fixed value as
$x$ tends to $\infty$. Then $f'(x)$ will become  positive after sometime and so $f(x)$ will be strictly increasing after that ( for similar reason $f(x)$ will not tend to a fixed value as x goes $\infty$) and hence $\lim\limits_{x \to \infty} f(x) =\infty$ .