Prove $\lim_{x\to\infty}\frac{f(x)}{x}=0$ if $f$ is continuous and $\lim_{x\to\infty}f(x+1)-f(x)=0$

Question

This is Problem 22-17 in Spivak's Calculus (3rd edition). It says to use the previous problem, which is if $\lim_{n\to\infty}a_n=l$, then $$\lim_{n\to\infty}\frac{\sum_{i=1}^n a_n}{n}=l.$$ I think I'm supposed to let $a_n=f(n)-f(n-1)$ and then $\sum_{i=1}^n a_n=f(n)-f(0)$, so $$\frac{\sum_{i=1}^n a_n}{n}=\frac{f(n)-f(0)}{n}$$ which has the same limiting behavior as $\frac{f(n)}{n}$. But I can't figure out how to use the continuity of $f$, as something like a rescaled version of $\tan$ is periodic and so $f(n+1)-f(n)=0$, but $\lim_{x\to\infty}\frac{\tan(x)}{x}\neq 0$.

Answer 1

Pick $\epsilon>0$ and $M>0$ such that if $x \ge M$ then $|f(x)-f(x+1)| < {1 \over 2}\epsilon$.

Let $L = \sup_{t \in [M,M+1]} |f(t)|$ (this is where continuity is used), then for $x\ge M$ we have $x = M + (x- \lfloor x-M\rfloor) +\lfloor x-M\rfloor $. Let $x_0 = M + (x- \lfloor x-M\rfloor)$ and note that $x_0 \in [M,M+1]$. Then \begin{eqnarray} f(x) &=& f( x_0 +\lfloor x-M\rfloor) \\ &=& f(x_0)+\sum_{k=1}^{\lfloor x-M\rfloor} (f( x_0 +k)-f( x_0 +k-1) \end{eqnarray}

Then $|f(x)| \le L + \lfloor x-M\rfloor {1 \over 2} \epsilon$ and so ${|f(x)| \over x} = {L \over x} + { \lfloor x-M\rfloor \over x} {1 \over 2} \epsilon \le {L \over x} + {1 \over 2} \epsilon$. Now chose $M' \ge M$ such that ${L \over x} < {1 \over 2} \epsilon$ and then we have ${|f(x)| \over x} < \epsilon $ for all $x > M'$.