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 $k > 0$. Prove that $\lim_{x \to \infty} f(x) = \infty$.

I did the following:
If $\lim_{x \to \infty} f''(x) = k > 0$, then this means that: $$\exists M\in \mathbb R: x > M \implies f''(x)>0$$
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 $>0$.
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 $x > 0$.
How can I prove this?
 A: $f''(x) > 0$ for large $x$ is not sufficient. But you have
$$
 f''(x) \ge \frac 12 k > 0
$$
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.
A: Since $ \lim\limits_{x\to +\infty}{f''\left(x\right)}=k $, there exists some $ A > 0 $ such that $ \left(\forall x> A\right),\ \left|f''\left(x\right)-k\right|<\frac{k}{2} $
If $ a $ is such that $ f'\left(a\right) $ exists, and $ f''\left(x\right) $ exists for all $ x>a $, then we have, for all $ x>A $ : \begin{aligned} f\left(x\right)&= f(a)+\left(x-a\right)f'\left(a\right)+\int_{a}^{x}{\left(x-y\right)f''\left(y\right)\mathrm{d}y}\\ &\geq f\left(a\right)+\left(x-a\right) f'\left(a\right)+\frac{k}{2}\int_{a}^{x}{\left(x-y\right)\mathrm{d}y}\\ &\geq f\left(a\right)+\left(x-a\right)f'\left(a\right)+\frac{k\left(x-a\right)^{2}}{4}\underset{x\to +\infty}{\longrightarrow}+\infty \end{aligned}
Thus : $$ \lim_{x\to +\infty}{f\left(x\right)}=+\infty $$
A: For $x$ large enough let say $x \ge X$, you have
$$f^{\prime \prime} (x) \ge \frac{k}{2} \gt 0$$ and by Mean Value Theorem
$$f^\prime(x) - f^\prime(X) \ge \frac{k}{2}(x-X).$$
This implies that $\lim\limits_{x \to \infty} f^{\prime}(x) = \infty$. In particular $f^\prime(x) \gt K \gt 0$ for $x$ large enough.
Apply what we did above to $f^{\prime \prime}$ to $f^\prime$ and you're done.
A: 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 $\lim\limits_{x \to \infty} f''(x) > 0$ 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 $\lim\limits_{x \to \infty} f(x) =\infty$ .
