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.

For $t,x\in\mathbb R$, let $h_t(x) = f(t+x)-f(t-x)$.
For $a,t,x\in\mathbb R$, let $g_{a,t}(x)=h_{x+t}(2a)$.
Then $$\begin{align}g_{a,t}(x)&=f(x+2a+t)-f(x-2a+t)\\&=2(f(a+t)-f(t-a))\\&\quad{}+[f(x+2a+t)-2f(a+t)+f(t-x)]\\&\quad{}-[f(x-2a+t)-2f(t-a)+f(t-x)]\\
&\to2(f(a+t)-f(t-a))= 2h_t(a)&\text{as }x\to\pm\infty.\end{align} $$
Because $g_{a,t}(x)=g_{a,0}(x+t)$, we conclude that the limit does not depend on $t$, i.e. $h_t(a)=h_0(a)$. 
For $a,b\in\mathbb R$, we have
$$\begin{align}g_{a,0}(x)+g_{b,0}(-x)&=f(x+2a)-f(x-2a)+f(-x+2b)-f(-x-2b)\\
&=2f(a+b)-2f(-a-b)\\&\quad{}+[f(x+2a)-2f(a+b)+f(-x+2b)]\\&\quad{}-[f(x-2a)-2f(-a-b)+f(-x-2b)]\\&\to 2h_0(a+b)&\text{as }x\to\pm\infty. \end{align}$$
Thus $h_0(a+b)=h_0(a)+h_0(b)$.
Since $h_0$ is continuous, this implies that $h_0(x)=\alpha x$ for some $\alpha\in\mathbb R$. (Without continuity, $h_0$ might be an ugly endomorphism of the $\mathbb Q$-vector space $\mathbb R$).
We conclude 
$$ f(x) = f(0)+2h_{\frac x2}\bigl(\tfrac x2\bigr) = f(0)+2h_0\bigl(\tfrac x2\bigr)=f(0)+\alpha x.$$
