Cauchy Problem - Real Analysis Been struggling with this for a while now, just kind of spinning my wheels at this point.  Any help/advice/anything much appreciated.  This is a problem from Wade's Intro to Analysis, chapter 3.
Suppose $f : \textbf{N} \to \textbf{R}$. If $\lim\limits_{n \to \infty}f(n + 1) - f(n) = L$, prove that $\lim\limits_{n \to \infty}\frac{f(n)}{n}$ exists and equals $L$.

Suppose that we  are given an $\epsilon \gt 0$. Then there is a $b=b(\epsilon)$ such that if $k\ge b$, then $f(k+1)-f(k)$ is within $\epsilon/3$ of $L$. This $b$ will be fixed from now on. 
Note that
$$(f(b+1-f(b))+(f(b+2)-f(b+1))+\cdots +(f(n)-f(n-1))=f(n)-f(b).$$
On the left side we have $n-b$ terms whose sizes are under control Thus 
$$(n-b)(L-\epsilon/3) \lt f(n)-f(b) \lt (n-b)(L+\epsilon/3).$$
Add $f(b)$ to each of the three sides, and divide by $n$. We get
$$\frac{n-b}{n}(L-\epsilon/3)+\frac{f(b)}{n}\lt \frac{f(n)}{n} \lt \frac{n-b}{n}(L+\epsilon/3)+\frac{f(b)}{n}.$$
We conclude that for large enough $n$, $\frac{f(n)}{n}$ is within $\epsilon$ of $L$. It suffices to make sure that $\frac{f(b)}{n}$ is within $\epsilon/3$ of $0$, and that the term $\frac{n-b}{n}$ is close enough to $1$. 
