Prove that if $f$ is continuous on $\mathbb R$ and

$$\lim_{x \to +\infty} [f(x+1)-f(x)] = l,$$


$$\lim_{x\to +\infty} f(x)/x =l.$$

So I've been trying for hours to use the series definition of limits / Cesaro's lemma (if $\lim U(n+1) - U(n) = l$ then $\lim U(n)/n=l$).

I'm completly blocked and I can't get it.

  • $\begingroup$ Hi, I have edited parts of the question so that it is easier to read. But I am not sure what you want to say (for the last two/three lines), so I will just leave that to you. $\endgroup$
    – user99914
    Nov 27, 2014 at 23:01

1 Answer 1


Denoting, $\displaystyle U_n = \sup\limits_{x \in [n,n+1)} f(x)$ and $\displaystyle u_n = \inf\limits_{x \in [n,n+1)} f(x)$ (both sequences are well defined since, $f$ is continuous)

Since, for each $\epsilon > 0$, there is $x_n \in [n,n+1)$, such that $f(x_n) > U_n - \epsilon$.

Hence, $\displaystyle f(x_n+1) - f(x_n) -\epsilon \le U_{n+1} - U_n \le f(x_{n+1}) - f(x_{n+1} - 1) + \epsilon$, for each $n$.

Thus, $\lim\limits_{x \to \infty} f(x+1) - f(x) = \lim\limits_{n \to \infty} U_{n+1} - U_n = l$

Similarly, one can show that $\lim\limits_{n \to \infty} u_{n+1} - u_n = l$

Now a direct application of Stolz–Cesàro theorem, yeilds

$\displaystyle \lim\limits_{n \to \infty} \dfrac{U_n}{n} = \lim\limits_{n \to \infty} U_{n+1} - U_n = l$ and, $\lim\limits_{n \to \infty} \dfrac{u_n}{n+1} = l$.

Since, $\displaystyle \frac{u_n}{n+1} \le \dfrac{f(x)}{x} \le \dfrac{U_n}{n}$ for all $x \in [n,n+1)$, we conclude $\displaystyle \lim\limits_{x \to \infty} \frac{f(x)}{x} = l$.

  • 1
    $\begingroup$ Can you give some quick details of how you found $U_{n+1} - U_n \le f(x_{n+1}) - f(x_{n+1} - 1) + \epsilon$ ? $\endgroup$
    – Chady
    Jan 2, 2021 at 21:06
  • $\begingroup$ @Chady add the two inequalities $U_{n+1} < f(x_{n+1}) + \epsilon$ and $f(x_{n+1} - 1) \le U_n$. $\endgroup$
    – sciona
    Jan 9, 2021 at 7:00

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