Prove $f(x)=ax+b$ Let $f(x)$ be a continuous function in $\mathbb R$ that for all $x\in(-\infty,+\infty)$, satisfies
$$ \lim_{h\rightarrow+\infty}{[f(x+h)-2f(x)+f(x-h)]}=0. $$ 
Prove that $f(x)=ax+b$ for some $a,b\in\mathbb R$.
This is a problem from my exercise book, but I can't figure out the solution of it, I think the solution in my book is wrong. :( Any idea and proof of it are welcome! Thank you in advance.

We can assume that $f(-1)=0$ and $f(1)=0$ (because if not then replace $f(x)$ by $f(x)-\frac{f(1)-f(-1)}{2}(x+1) -f(-1)$). We prove that $f=0$, hence for general $f$ we will have $f(x) = \frac{f(1)-f(-1)}{2}(x+1) -f(-1)$.
The hypothesis for $x=1$ and $x=-1$ gives :
$$f(h)=-f(2-h)+o(1),$$
$$f(h)=-f(-2-h)=o(1),$$
when $h$ goes to $+\infty$.
Lemma : $f$ is periodic.
Proof. Take $x \in \mathbb{R}$. Then :
\begin{eqnarray*}
f(x+h)+f(x-h)
&=& 2 f(x) + o(1) \\
&=& -f(2-x-h) - f(2-x+h) + o(1) = -2f(2-x) +o(1) \\
&=& -f(2-x-h) - f(-2-x+h) + o(1) = -2f(-x) +o(1) \\
&=& -f(-2-x-h) - f(-2-x+h) + o(1) = -2f(-2-x) +o(1) \\
\end{eqnarray*}
Hence $f(x)=-f(-x)=-f(2-x)=-f(-2-x)$. This implies that $f$ is $2$-periodic. QED
Let $T$ be the group of periods of $f$ and $t \in T$. Then
\begin{eqnarray*}
\forall x, \ f(x+h)+f(x-h)
&=& 2 f(x) + o(1) \\
&=& f(x+h)+f(t+x-h) = 2 f(t/2+x) +o(1) \\
\end{eqnarray*}
Hence $\frac{t}{2} \in T$. So $T$ is non discrete and since $f$ is continuous, $T=\mathbb{R}$. 
