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

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.

• Take a look here $\,$
– r9m
Dec 16, 2014 at 22:08
• Maybe I'm not seeing it right, but I don't think that it applies here. For example, the function $f(x)=\frac{sin(x^2)}{x}$ has as asymptote the $x$ axis and $lim_{x\rightarrow\infty}f'(x)$ does not exit, even though $f'$ is bounded. Thanks for answering! Dec 16, 2014 at 22:17

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$be arbitrary, and in the original proof  set$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.