$f(x+1)-f(x)$ converges $\Rightarrow\frac{f(x)}x$ converges

Here is a similar but more interesting statement.

Let $f$ defined on $(a,+\infty)$ and bounded on all bounded interval $(a,b)$. If $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ then $\lim_{x\to\infty}\frac{f(x)}{x}=l$ as well.

Let $\lim_{x\to\infty}\bigr(f(x+1)-f(x)\bigl)=l$ and introduce $M_n=\sup_{[n,n+1)} f(x)$ and $m_n=\inf_{[n,n+1)} f(x)$. The sequences $\{M_n\}$ and $\{m_n\}$ are well defined for $n\ge \lfloor a\rfloor +1$. By definition of $\sup$ and $\inf$ we have have a sequence $\{x_n\}\in[n,n+1)$ sand $f(x_n)>M_n-\varepsilon$. Now conclude that $\lim_{n\rightarrow +\infty} (M_{n+1}-M_n)=l$ and $\lim_{n\rightarrow +\infty} (m_{n+1}-m_n)=l$ as well.

By Stolz-Cesaro theorem $$\lim_{n\rightarrow +\infty}\frac{M_n}{n}=\lim_{n\rightarrow +\infty}\frac{m_n}{n+1}.$$ Then there exist $n_0$ such that $n>n_0$ we have $$-\varepsilon<\frac{M_n}{n}-l<\varepsilon\quad\text{and}\quad -\varepsilon<\frac{m_n}{n+1}-l<\varepsilon.$$ It follows that $f(x)>0$ for $x$ large enough if $l>0$. Then if $n_x=\lfloor x\rfloor$ then $$ \frac{m_{n_x}}{n_x+1}\le\frac{f(x)}{x}\le\frac{M_{n_x}}{n_x} .$$ It's pretty easy to conclude for $l\ne 0$. I leave the rest of the job to you for $l=0$ and for $l<0$ the preceding inequality becomes $$ \frac{m_{n_x}}{n_x}\le\frac{f(x)}{x}\le\frac{M_{n_x}}{n_x+1} .$$

It seems the following.

Let $1>\varepsilon>0$ be an arbitrary number. There exists a number $N>0$ such that $$|f(x+1)-f(x)-l|\le \varepsilon$$ for each $x\ge N$. Since the function $f$ is continuous, $$\sup \{f(y): y\in [N;N+1]\}=M<\infty.$$ Let $y\ge \max \{M, (|l|+1)(N+1)\}/\varepsilon $ be an arbitrary number. There exists a nonegative integer $k_y$ such that $N\le y-k_y<N+1$. Then $k_y\le y$, $M\le y\varepsilon$, and $|l|(y-k_y)<|l|(N+1)\le y\varepsilon$. So

$$|f(y)-ly|\le$$ $$|f(y)-f(y-1)-l|+|f(y-1)-f(y-2)-l|+\dots+|f(y-k_y+1)-f(y-k_y)-l|+|f(y-k_y)|+|lk_y-ly|\le k_y\varepsilon+M+|l|(y-k_y)\le 3y\varepsilon. $$