If a function has asymptote and the derivative does not, then its second derivative is not bounded

Question

Let $f\colon\mathbb{R} \rightarrow \mathbb{R}$ be a function with second derivative everywhere in is domain. Prove that if $\lim_{x\rightarrow\infty}f(x)=b \in \mathbb{R}$ and $\lim_{x\rightarrow\infty}f'(x)$ does not exist (and it is not infinity either!), then $f''(x)$ cannot be bounded.

Answer 1

As $r9m$ suggested above, you can adapt the second proof given by Giuseppe Negro on this thread as follows: by contradiction, suppose $f$ has a limit $b$ at $\infty$, $f^\prime$ has no limit at infinity, but $f^{\prime\prime}$ is bounded$.

First, without loss of generality one can assume $b=0$. Then, let $\varepsilon > 0$ be arbitrary, and in the original proof [1] set $x_\varepsilon$ be such that $\forall x \geq x_\varepsilon$, $\lvert f(x)\rvert \leq \varepsilon^2$; and take any $x_0 \geq x^\ast$. Then, instead of $C_0$ you can write $\varepsilon^2$ in the proof, yielding the conclusion $$\lvert f'(x_0)\rvert \leq 2\sqrt{2} \sqrt{\varepsilon^2C_2} = \varepsilon\cdot2\sqrt{2} \sqrt{C_2}.$$ As $\varepsilon$ was arbitrary and $x_0$ is an arbitrary element greater than $x_\varepsilon$, you get $f^\prime(x) \xrightarrow[x\to\infty]{} 0$, hence a contradiction.