Limited function.

Question

Let $f:[0,+\infty) \rightarrow \mathbb{R}$ be a function bounded on each bounded interval. Prove that if $\lim_{x \rightarrow +\infty } [f(x+1)-f(x)] = L$, then $\lim_{x \rightarrow +\infty} \frac{f(x)}{x} = L$.

Answer 1

Assume $|f(x+1)-f(x)-L|<\epsilon$ for $x \ge M$.

Use the triangle inequality as follows, for $x \ge M$:

\begin{align*} |f(x+n)- f(x)-nL| &= \left\rvert \sum_{i = 1}^n f(x+i)-f(x+i-1)-L \right\rvert \\ &\le \sum_{i=1}^n |f(x+i)-f(x+i-1)-L| \\ & < \sum_{i=1}^n \epsilon \\ & = n\epsilon \end{align*}

This gives you $|f(x+n)-f(x)-nL|<n\epsilon$. Now divide by $x+n$ and see what you get in the limit as $n \to \infty$, $x$ fixed. You should see that the left hand side of the inequality apporaches $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert$, while the right hand side approaches $\epsilon$.

Exercise: Suppose $|f(x)|<B$. In terms of $x$, $B$, and $L$, find a natural number $N$ large enough so that, for $n>N$, $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert<2\epsilon$. To do this, note that by the triangle inequality: $$\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert \le \left\lvert\left(\frac{f(x+n)}{x+n}-L\right) - \left(\frac{f(x+n)-f(x)-nL}{x+n}\right)\right\rvert + \left\lvert\frac{f(x+n)-f(x)-nL}{x+n}\right\rvert.$$ The quantity $\left\lvert\frac{f(x+n)-f(x)-nL}{x+n}\right\rvert$ is already bounded by $\epsilon$; you just need to simplify $\left\lvert\left(\frac{f(x+n)}{x+n}-L\right) - \left(\frac{f(x+n)-f(x)-nL}{x+n}\right)\right\rvert$ and show that $N$ exists so that it is bounded by $\epsilon$ as well.

Now consider $x \in [M,M+1]$. Use the fact that $f$ is bounded on $[M,M+1]$, say by $B$, and the fact that $M \le x \le M+1$, and your work above, to get the bound $\left\lvert\frac{f(x+n)}{x+n}-L\right\rvert<2\epsilon$ for $n>N$, for some choice of $N$. By substitution, we get the result that $\left\lvert \frac{f(x)}{x} - L \right\rvert < 2\epsilon$ for $x>M+N$. Now take $\epsilon$ to zero.

Answer 2

I originally misread the assumption as $\lim_{x\to\infty}|f(x+1)-f(x)|=L$. The result is false in that case:

Recall that the integer part $\lfloor x\rfloor$ of the real $x$ is the largest integer $n$ less than or equal to $x$, so for example $\lfloor-1.35\rfloor=-2$, $\lfloor e\rfloor=2$. Consider the function $f(x)$ that takes the value $1$ if the integer part of $x$ is even, and the value $0$ if the integer part of $x$ is odd. For example, $f(2.178)=f(-1.2)=1$ and $f(-3)=f(\pi)=0$. We have that $|f(x+1)-f(x)|=1$ for all $x$, and yet $\displaystyle \lim_{x\to\infty}\frac{f(x)}x=0$.

For the problem as stated, note that the assumption that $f$ is bounded on bounded intervals cannot be eliminated. An example illustrating this is Conway's base 13 function, that takes every value on any interval. Now let $f$ on $[0,1)$ coincide with Conway's function, and define $f(t+n)=f(t)+n$ for $n\in\mathbb Z$ and $t\in[0,1)$, so $f(x+1)-f(x)=1$ for all $x$. However, $\displaystyle \lim_{x\to\infty}\frac{f(x)}x$ does not exist.

Anyway, Dustan's argument gives the result under the given assumptions.

(Under the stronger assumption that $\lim_{x\to\infty}f'(x)=L$, an easier proof follows from L'Hôpital's rule. To see that this assumption implies the given one, note that if $f$ is differentiable, then $f(x+1)-f(x)=f'(t)$ for some $t\in(x,x+1)$, by the mean value theorem.)