\documentclass[a4paper,12pt]{article}
\begin{document}

\noindent {\bf Question}

\noindent Prove, if $f$ is continuous and if
$\lim_{x\rightarrow\infty} (f(x+1)-f(x)) =0$, that
$$\lim_{x\rightarrow\infty} \frac{f(x)}{x} =0.$$


\medskip

\noindent {\bf Answer}

\noindent First, since $\lim_{x\rightarrow \infty} (f(x+1) -f(x))
=0$, for any $\varepsilon
>0$, there exists $x_0$ (which we can take to be positive) so that
$|f(x+1) -f(x)| <\frac{1}{2} \varepsilon$ for $x >x_0$.  Now,
using the maximum value property (see note below), there exists a
maximum value $M$ of $|f(x)|$ on the interval $[x_0, x_0+1]$.

\medskip
\noindent The first claim is that for any $k\ge 0$, we have that
$|f(x)|\le \frac{k}{2} k\varepsilon +M$ for $x$ in the interval
$[x_0+k, x_0+k+1]$.  To see this, let $K$ be the maximum value of
$|f(x)|$ on $[x_0+k, x_0+k+1]$, occurring at $y$. Then, $x_0 +k\le
y\le x_0+k+1$, and so $x_0\le y -k\le x_0+1$. We now engage in
some algebraic manipulation:
\begin{eqnarray*}
|f(y)| & = &  |f(y) - f(y-k) + f(y-k)| \\
 & \le &  |f(y) - f(y-k)| + |f(y-k)| \\
 & \le &  |f(y) - f(y-1) + f(y-1) -\cdots f(y-k+1) + f(y-k+1) -
 f(y-k)| + |f(y-k)| \\
 & \le &  |f(y) - f(y-1)| + |f(y-1) - f(y-2)| + \cdots + |f(y-k+1) -
 f(y-k)| + |f(y-k)| \\
 & \le & \frac{1}{2} \varepsilon + \frac{1}{2} \varepsilon + \cdots +
 \frac{1}{2} \varepsilon + M \\
 & \le & \frac{k}{2} \varepsilon + M.
\end{eqnarray*}
In particular, this tells us that
\[ \frac{|f(y)|}{y}\le \frac{\frac{k}{2} \varepsilon +M}{y}\le
\frac{\frac{k}{2} \varepsilon}{y} +\frac{M}{y} \le
\frac{\frac{k}{2} \varepsilon}{x_0+k} +\frac{M}{y} <
\frac{\frac{k}{2} \varepsilon}{k} +\frac{M}{y} < \frac{1}{2}
\varepsilon +\frac{M}{y} \] for all $y$ in the interval $[x_0 +k,
x_0+k+1]$.

\medskip
\noindent Now, choose $x_1 > x_0$ so that $\frac{M}{x_1} <
\frac{1}{2} \varepsilon$.  Then, for all $y >x_1$, we have that
\[ \left| \frac{f(y)}{y}\right| = \frac{|f(y)|}{y} < \frac{1}{2}
\varepsilon +\frac{M}{y} < \frac{1}{2}\varepsilon +
\frac{1}{2}\varepsilon \] for all $y >x_1$.  In particular, we
have that the definition of $\lim_{x\rightarrow\infty}
\frac{f(x)}{x} =0$ is satisfied, as desired.


\noindent Note:

\noindent {\bf Maximum value property for continuous functions:}
Let $f$ be a function that is continuous on the closed interval
$[a,b]$. Then $f$ achieves its maximum on $[a,b]$; that is, there
exists some $x_0$ in $[a,b]$ so that $f(x_0)\ge f(x)$ for all
$x\in [a,b]$.
\end{document}