Prove that $\lim_{x \to +\infty} \frac{f(x)}{x} = L$ if $\lim_{x \to +\infty} [f(x+1) - f(x)] = L \space$ Let $f:[0, +\infty) \rightarrow \mathbb{R} $ be a bounded function in each bounded interval. If $$\lim_{x \to +\infty} [f(x+1) - f(x)] = L$$ then $$\lim_{x \to +\infty} \frac{f(x)}{x} = L$$ 
I tried using the definition on the first limit; then, I attempted to use that inequality to apply the triangular inequality to arrive at the definition that gives me the second limit. It is supposed to be simple, but I'm not feeling safe as in how I should write it.
 A: Using the condition, for all $\epsilon >0$, there is $N=N_\epsilon$ large so that 
$$ |f(x+1) - f(x) -L|< \epsilon$$
for all $x\ge N$. Then for each $M = 1, 2, 3, \cdots$, by induction on $M$ we have 
$$ | f(x+ M) - f(x) -ML|< M\epsilon$$
if $x\ge N$. As $f$ is bounded on bounded intervals, there is $K >0$ so that $|f(x) | \le K$ when $x\in [N, N+1]$. So for all $x\in [N, N+1]$, $M\in \mathbb N$ we have 
$$ |f(x+M)-ML| < M\epsilon +K.$$
Now for all $y\ge N$, there is $M$ so that $y=x+M$ for some $x\in [N, N+1]$. So for all $y\ge N$, 
$$\begin{split}
|f(y) - yL| &= |f( x+M) - ML + ML -yL| \\
&\le |f(x+M) - ML| + |(M-y)L| \\
&< M\epsilon +K +(N+1)|L| \\
&\le (y -N)\epsilon + K +(N+1)|L| \\
&= y\epsilon +C,
\end{split}$$
where $C = -N\epsilon + K +(N+1)|L|$ is independent of $y$. So 
$$ \left|\frac{f(y)}{y}  -L\right| \le \epsilon + \frac{C}{y}.$$
Now choose $N_1 \ge N$ so that $\frac{C}{y} <\epsilon$ whenever $y\ge N_1$. So 
$$\left|\frac{f(y)}{y}  -L\right| \le 2\epsilon$$
whenever $y\ge N_1$. Thus 
$$\lim_{x\to\infty} \frac{f(x)}{x} = L.$$
A: For each $\epsilon > 0$, there exists $N > 0$, such that for $x > N$, we have 
$$ |f(x + 1) - f(x) - L | < \epsilon$$.
Expand it, we have
$$ L -\epsilon + f(x) < f(x + 1) < f(x) + L + \epsilon$$
Use above relation, for each $y > N+ 1$, we have for some  $  N \le z \le N + 1 $,
$$  f(z) - z (L - \epsilon) < f(y) < f(z) +(y - z) (L + \epsilon) $$
Rewrite the abouve as
$$  f(z) + (- z) (L - \epsilon) < f(y) - yL < f(z) - z (L + \epsilon) $$
Furthermore
$$ |f(y) - Ly| \le \max_{N\le z\le N+1}|f(z)| + (N+1)(L+\epsilon)$$
Now,
let $M > N + 1 >0$ so that $\frac{max_{N<Z<N+1}|f(z)| + (N+1)(L+ \epsilon)}{M} \le \epsilon$
for $x > M$, $$|\frac{f(x)}{x} - L |= |\frac{f(x) - Lx}{x}|$$
$$ \le \frac {\max_{N\le z\le N+1}|f(z)| + (N+1)(L+\epsilon)}{M}$$
$$\le \epsilon$$
QED
