for $f$ continous, if $\lim_{x \to \infty} f(x+1)-f(x) = 0$ then $\lim_{x\to \infty}f(x)/x = 0$ (using sequences)

Question

The question is: for a continous function $f : \mathbb R \to \mathbb R,$ if $\lim_{x \to \infty} f(x+1)-f(x) = 0,$ prove $\lim_{x\to \infty}f(x)/x = 0$.

Now I was able to do it using only continuity and infinite limit definitions, but the question asks to use the sequences $a_n = \inf\limits_{[n,n+1]} f$ and $ b_n = \sup\limits_{[n,n+1]} f.$

I believe the result will follow if I can show $\lim_{n \to \infty} a_{n+1} - a_n = 0,$ and similarly for $b_n$, and then using a Cesaro sum related result (because the first part of the problem is to show $a_{n}/n \to l$ if $a_{n+1} - a_n \to l$ ), but I'm completely stuck proving this last limit.

Edit: Thanks for the help. While the linked answer solves the question in a broader sense, it doesn't use the sequences the specific problem asks to use (this question is in a chapter about sequences, so I have to fit it here somehow).

Now I believe I've solved the problem. Using the defined $b_n,$ for each $n \in \mathbb N,$ there is a $x_n \in [n,n+1]$ such that $b_n - 1/n < f(x_n)$ (definition of sup with $\epsilon = 1/n$) and also $f(x_{n} +1) \leq b_{n+1},$ and similar inequalities for $b_{n+1}.$ With this, $$f(x_n +1) - f(x_n) -\tfrac{1}{n} < b_{n+1} - b_{n} < f(x_{n+1}) - f(x_{n+1} -1) +\tfrac{1}{n+1} $$ so $b_{n+1} - b_n \to 0,$ and similarly $a_{n+1} - a_n \to 0.$ A corollary of Stolz-Cesaro implies $a_n/n , b_n/n \to 0,$ and in particular $\frac{a_n}{n+1} \to 0 $ and $\frac{b_n}{n+1} \to 0$ Therefore, given $\epsilon> 0 ,$ there is a $N \in \mathbb N$ large enough such that for $n > N,$ $|a_n/n| < \epsilon,$ $|b_n/n| < \epsilon,$ $|\frac{a_n}{n+1}| < \epsilon$ and $|\frac{b_n}{n+1}| < \epsilon.$ So for $x > N,$ $x \in [n,n+1]$ for some natural $n > N$ and we have, if $f(x)\geq 0,$ $b_n \geq 0$ and
$$ -\epsilon < \frac{a_n}{n+1}\leq \frac{f(x)}{x} \leq \frac{b_n}{n}< \epsilon,$$ and if $f(x) \leq 0,$ $a_n \leq 0,$ and we use the same reasoning as above. In any case, $|f(x)/x| < \epsilon$ if $x > N.$

Answer 1

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. Suppose that $\lim_{x\rightarrow\infty}f(x+1)-f(x)=0$, then $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=0$.

Proof: Let $\varepsilon>0$. Then there exists $X_{0}>0$ such that $|f(x+1)-f(x)|<\varepsilon$ whenever $x\geq X_{0}$. By triangular inequality, for any $x\in[X_{0},X_{0}+1)$, $k\in\mathbb{N}$, we have $|f(x+k)-f(x)|<k\varepsilon$. Let $M=\sup_{x\in[X_{0},X_{0}+1]}|f(x)|$. By continuity of $f$ and compactness of $[X_{0},X_{0}+1]$, $M<\infty$. Choose $Y_{0}>X_{0}$ such that $\frac{M}{Y_{0}}<\varepsilon$.

For each $y\in(Y_{0},\infty)$, there exist uniquely $x\in[X_{0},X_{0}+1)$ and $k\in\mathbb{N}\cup{\{0\}}$ such that $y=x+k$. Now \begin{eqnarray*} \left|\frac{f(y)}{y}\right| & \leq & \left|\frac{f(y)-f(x)}{y}\right|+\left|\frac{f(x)}{y}\right|\\ & \leq & \frac{k\varepsilon}{y}+\frac{M}{Y_{0}}\\ & \leq & \varepsilon+\varepsilon\\ & = & 2\varepsilon. \end{eqnarray*} This shows that $\lim_{x\rightarrow\infty}\frac{f(x)}{x}=0$.

Note that the above proof remains valid even $f$ is not continuous, provided that there exist sufficiently large $a$ and $b$, with $a<b$ and $b-a\geq1$ such that $\sup_{x\in[a,b)}|f(x)|<\infty.$