Does $\lim_{x\to\infty}\frac{f(x) - f(x/2)}{x} = \ell$ say anything about the convergence of $\lim_{x\to\infty} \frac{f(x)}{x}$? Title pretty much says all.
Of course, if $\lim_{x\to\infty}\frac{f(x)}{x}$ exists and is equal to $a$, then
\begin{eqnarray*}
\ell &=& \lim_{x\to\infty} \frac{f(x) - f(x/2)}{x} \\
&=& \lim_{x\to\infty} \frac{f(x)}{x} -\lim_{x\to\infty} \frac{1}{2} \frac{f(x/2)
}{x/2} \\
&=& a - a/2 = a/2.
\end{eqnarray*}
That is, $a = 2\ell$. Does the existence of the limit $\ell$ ensure the existence of the limit $a$?
 A: The answer is not unless you add extra conditions on $f$. 
For example, define on $(0, \infty)$ the following equivalence relation:
$$x \sim y \Leftrightarrow \frac{x}{y} \in \mathbb Q$$
For this relation you can pick a system $B$ of representatives which is unbounded.
Now, for each $b \in B$ define 
$$f(x)=b \mbox { whenever } x \sim b$$
For this function you always have $f(x)=f(x/2)$.
But since $B$ is unbounded, you can find a sequence $b_n \in B$ such that $b_n \to \infty$.
Then
$$b_n \to \infty \mbox{ and } \lim_n \frac{f(b_n)}{b_n}=1$$
while
$$n \to \infty   \mbox{ and } \lim_n \frac{f(n)}{n}= \lim_n \frac{f(1)}{n}=0$$
To eliminate this counterexample, you need to assume that $f$ is at least continuous, maybe differentiable would bea good extra condition.
A: Theorem 1:
If $\;g:\left[0,+\infty\right[\to\mathbb{R}\;$ is a continuous function on $\left[0,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
Proof:
Let $\;\beta=
  \begin{cases}
    \alpha & \text{if }\;0<\alpha<1 \\
    \cfrac{1}{\alpha} & \text{if }\;\alpha>1
  \end{cases}\;.\;\;$ It results that $\;0<\beta<1.$
We are going to prove that
$\cfrac{1}{1-\beta}\lim\limits_{x\to+\infty}\big[g(x)-\beta g(\beta x)\big]=\cfrac{l}{1-\alpha}\;.\quad\color{blue}{(*)}$
The previous equality is evidently true if $\;\beta=\alpha\;.$
If $\;\beta=\frac{1}{\alpha}\;,\;$ it follows that
$\frac{1}{1-\beta}\lim\limits_{x\to+\infty}\big[g(x)-\beta g(\beta x)\big]=$
$=\frac{\alpha}{\alpha-1}\lim\limits_{x\to+\infty}\big[g(x)-\frac{1}{\alpha} g\left(\frac{x}{\alpha}\right)\big]=$
$=\frac{1}{\alpha-1}\lim\limits_{x\to+\infty}\big[\alpha g(x)-g\left(\frac{x}{\alpha}\right)\big]=$
$=\frac{1}{1-\alpha}\lim\limits_{x\to+\infty}\big[g\left(\frac{x}{\alpha}\right)-\alpha g(x)\big]=$
$\underset{\overbrace{\text{ by letting } X=\frac{x}{\alpha}\;}}{=}\frac{1}{1-\alpha}\lim\limits_{X\to+\infty}\big[g(X)-\alpha g(\alpha X)\big]=\frac{l}{1-\alpha}\;.$
Moreover,
$\sum\limits_{n=0}^{k}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=g(x)-\beta^{k+1} g\left(\beta^{k+1} x\right)$
for all $\;x\in\left[0,+\infty\right[\;$ and for all $\;k\in\mathbb{N}=\{0,1,2,3,\ldots\}\;.$
For any $\;x\in\left[0,+\infty\right[\;$, since $\;\lim\limits_{k\to\infty}\left(\beta^{k+1}x\right)=0\;$ and the function $\;g\;$ is continuous at the point $x_0=0\;,\;$ we get that $\lim\limits_{k\to\infty}g\left(\beta^{k+1}x\right)=g(0)\;,\;$ consequently,
$\lim\limits_{k\to\infty}\big[g(x)-\beta^{k+1} g\left(\beta^{k+1} x\right)\big]=g(x)\;,\;\;\;\;\forall\;x\in\left[0,+\infty\right[.$
Therefore,
$g(x)=\sum\limits_{n=0}^{\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]\quad\color{blue}{(**)}$
for all $\;x\in\left[0,+\infty\right[\;.$
Since the function $\;g(x)-\beta g(\beta x)\;$ is continuous on $\left[0,+\infty\right[$ and $\lim\limits_{x\to+\infty}\big[g(x)-\beta g(\beta x)\big]\underset{\overbrace{\text{ from }(*)\;}}{=}\cfrac{l(1-\beta)}{1-\alpha}\in\mathbb{R}\;,\;$
there exists $\;M>0\;$ such that
$\big|g(x)-\beta g(\beta x)\big|\le M\;,\quad$ for all $\;x\in\left[0,+\infty\right[\;,\;$ hence,
$\big|g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big|\le M\;,\;\;\forall x\in\left[0,+\infty\right[\;$ and $\;\forall n\in\mathbb{N}$.
So the sequence $\;\big\{\beta^n M\big\}_{n\in\mathbb{N}}\;$ of positive real numbers satisfies the following conditions:
$\bigg|\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]\bigg|\le \beta^n M$,
$\forall x\in\left[0,+\infty\right[\;$ and $\;\forall n\in\mathbb{N}\;,$
$\sum\limits_{n=0}^{\infty}\big(\beta^n M\big)=\cfrac{M}{1-\beta}\in\mathbb{R}^+\;.$
By applying the Weierstrass M-test, we get that the series $(**)$ is uniformly convergent on $\left[0,+\infty\right[,\;$ consequently
$\exists\lim\limits_{x\to+\infty}g(x)=\lim\limits_{x\to+\infty}\sum\limits_{n=0}^{\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=$
$\underset{\overbrace{\text{ for the uniform convergence }}}{=}\;\sum\limits_{n=0}^{\infty}\lim\limits_{x\to+\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=$
$\underset{\overbrace{\text{ by letting } X=\beta^n x\;}}{=}\;\sum\limits_{n=0}^{\infty}\lim\limits_{X\to+\infty}\beta^n\big[g(X)-\beta g(\beta X)\big]=$
$\underset{\overbrace{\text{ from }(*)\;}}{=}\;\sum\limits_{n=0}^{\infty}\beta^n\cfrac{l(1-\beta)}{1-\alpha}=\cfrac{l(1-\beta)}{1-\alpha}\sum\limits_{n=0}^{\infty}\beta^n=$
$=\cfrac{l(1-\beta)}{1-\alpha}\cdot\cfrac{1}{1-\beta}=\cfrac{l}{1-\alpha}\;.$

Corollary 1:
If $\;g:\left[a,+\infty\right[\to\mathbb{R}\;$ is a continuous function on $\left[a,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
Proof:
If $\;a\le0\;$ the restriction $\;g\big|_{\left[0,+\infty\right[}\;$ of the function $g$ to $\left[0,+\infty\right[$ satisfies all the hypotheses of the previous theorem, so there exists $\lim\limits_{x\to+\infty}g(x)=\lim\limits_{x\to+\infty}g\big|_{\left[0,+\infty\right[}=\cfrac{l}{1-\alpha}\;.$
If $\;a>0\;$ we consider the function $\;G:\left[0,+\infty\right[\to\mathbb{R}\;$ defined as
$G(x)= \begin{cases}
    g(a) & \text{if }\;0\le x<a\\
    g(x) & \text{if }\;x\ge a
  \end{cases}\;.$
The function $\;G\;$ satisfies all the hypotheses of the previous theorem, so there exists $\lim\limits_{x\to+\infty}g(x)= \lim\limits_{x\to+\infty}G(x)=\cfrac{l}{1-\alpha}\;.$

Corollary 2:
If $\;a\;$ is a positive real number $\;(a>0)\;,$
$f:\left[a,+\infty\right[\to\mathbb{R}\;$ is a continuous function on $\left[a,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\cfrac{f(x)-f(\alpha x)}{x}=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}\cfrac{f(x)}{x}=\cfrac{l}{1-\alpha}\;.$
Proof:
We consider the function $\;g:\left[a,+\infty\right[\to\mathbb{R}\;$ defined as
$g(x)=\cfrac{f(x)}{x}\quad$ for all $\;x\in\left[a,+\infty\right[\;.$
The function $\;g(x)\;$ is continuous on $\;\left[a,+\infty\right[\;$ and
$\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=\lim\limits_{x\to+\infty}\left[\cfrac{f(x)}{x}-\alpha\cfrac{f(\alpha x)}{\alpha x}\right]=$
$=\lim\limits_{x\to+\infty}\cfrac{f(x)-f(\alpha x)}{x}=l\in\mathbb{R}\;.$
The function $\;g(x)\;$ satisfies all the hypotheses of Corollary 1, so there exists $\lim\limits_{x\to+\infty}\cfrac{f(x)}{x}= \lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
A: $\color{blue}{\text{It is not necessary that the function } g \text{ is continuous,}}\\\color{blue}{\text{in fact it is sufficient that it is bounded.}}$
Theorem 2:
If $\;g:\left[0,+\infty\right[\to\mathbb{R}\;$ is a bounded function on $\left[0,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
Proof:
Let $\;\beta=
  \begin{cases}
    \alpha & \text{if }\;0<\alpha<1 \\
    \cfrac{1}{\alpha} & \text{if }\;\alpha>1
  \end{cases}\;.\;\;$ It results that $\;0<\beta<1.$
We are going to prove that
$\cfrac{1}{1-\beta}\lim\limits_{x\to+\infty}\big[g(x)-\beta g(\beta x)\big]=\cfrac{l}{1-\alpha}\;.\quad\color{blue}{(*)}$
The previous equality is evidently true if $\;\beta=\alpha\;.$
If $\;\beta=\frac{1}{\alpha}\;,\;$ it follows that
$\frac{1}{1-\beta}\lim\limits_{x\to+\infty}\big[g(x)-\beta g(\beta x)\big]=$
$=\frac{\alpha}{\alpha-1}\lim\limits_{x\to+\infty}\big[g(x)-\frac{1}{\alpha} g\left(\frac{x}{\alpha}\right)\big]=$
$=\frac{1}{\alpha-1}\lim\limits_{x\to+\infty}\big[\alpha g(x)-g\left(\frac{x}{\alpha}\right)\big]=$
$=\frac{1}{1-\alpha}\lim\limits_{x\to+\infty}\big[g\left(\frac{x}{\alpha}\right)-\alpha g(x)\big]=$
$\underset{\overbrace{\text{ by letting } X=\frac{x}{\alpha}\;}}{=}\frac{1}{1-\alpha}\lim\limits_{X\to+\infty}\big[g(X)-\alpha g(\alpha X)\big]=\frac{l}{1-\alpha}\;.$
Moreover,
$\sum\limits_{n=0}^{k}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=g(x)-\beta^{k+1} g\left(\beta^{k+1} x\right)$
for all $\;x\in\left[0,+\infty\right[\;$ and for all $\;k\in\mathbb{N}=\{0,1,2,3,\ldots\}\;.$
Given that the function $\;g\;$ is bounded on $\;\left[0,+\infty\right[,\;$ there exists $\;M>0\;$ such that
$\left|g(x)\right|\le M\;$ for all $\;x\in\left[0,+\infty\right[\;.$
Since $\;\lim\limits_{k\to\infty}\beta^{k+1}=0\;$ and $\;\left|g\left(\beta^{k+1} x\right)\right|\le M\;$ for all $\;x\in\left[0,+\infty\right[\;$ and for all $\;k\in\mathbb{N}\;,\;$ we get that
$\lim\limits_{k\to\infty}\big[g(x)-\beta^{k+1} g\left(\beta^{k+1} x\right)\big]=g(x)\;,\;\;\;\;\forall\;x\in\left[0,+\infty\right[.$
Therefore,
$g(x)=\sum\limits_{n=0}^{\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]\quad\color{blue}{(**)}$
for all $\;x\in\left[0,+\infty\right[\;.$
Moreover,
$\big|g(x)-\beta g(\beta x)\big|\le \big|g(x)\big|+\beta\big|g(\beta x)\big|\le M+\beta M=N\in\mathbb{R}^+\;,$
for all $\;x\in\left[0,+\infty\right[\;,\;$ hence,
$\big|g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big|\le N\;,\;\;\forall x\in\left[0,+\infty\right[\;$ and $\;\forall n\in\mathbb{N}$.
So the sequence $\;\big\{\beta^n N\big\}_{n\in\mathbb{N}}\;$ of positive real numbers satisfies the following conditions:
$\bigg|\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]\bigg|\le \beta^n N$,
$\forall x\in\left[0,+\infty\right[\;$ and $\;\forall n\in\mathbb{N}\;,$
$\sum\limits_{n=0}^{\infty}\big(\beta^n N\big)=\cfrac{N}{1-\beta}\in\mathbb{R}^+\;.$
By applying the Weierstrass M-test, we get that the series $(**)$ is uniformly convergent on $\left[0,+\infty\right[,\;$ consequently
$\exists\lim\limits_{x\to+\infty}g(x)=\lim\limits_{x\to+\infty}\sum\limits_{n=0}^{\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=$
$\underset{\overbrace{\text{ for the uniform convergence }}}{=}\;\sum\limits_{n=0}^{\infty}\lim\limits_{x\to+\infty}\beta^n\big[g\left(\beta^n x\right)-\beta g\left(\beta^{n+1} x\right)\big]=$
$\underset{\overbrace{\text{ by letting } X=\beta^n x\;}}{=}\;\sum\limits_{n=0}^{\infty}\lim\limits_{X\to+\infty}\beta^n\big[g(X)-\beta g(\beta X)\big]=$
$\underset{\overbrace{\text{ from }(*)\;}}{=}\;\sum\limits_{n=0}^{\infty}\beta^n\cfrac{l(1-\beta)}{1-\alpha}=\cfrac{l(1-\beta)}{1-\alpha}\sum\limits_{n=0}^{\infty}\beta^n=$
$=\cfrac{l(1-\beta)}{1-\alpha}\cdot\cfrac{1}{1-\beta}=\cfrac{l}{1-\alpha}\;.$

Corollary 3:
If $\;g:\left[a,+\infty\right[\to\mathbb{R}\;$ is a bounded function on $\left[a,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
Proof:
If $\;a\le0\;$ the restriction $\;g\big|_{\left[0,+\infty\right[}\;$ of the function $g$ to $\left[0,+\infty\right[$ satisfies all the hypotheses of the previous theorem, so there exists $\lim\limits_{x\to+\infty}g(x)=\lim\limits_{x\to+\infty}g\big|_{\left[0,+\infty\right[}=\cfrac{l}{1-\alpha}\;.$
If $\;a>0\;$ we consider the function $\;G:\left[0,+\infty\right[\to\mathbb{R}\;$ defined as
$G(x)= \begin{cases}
    g(a) & \text{if }\;0\le x<a\\
    g(x) & \text{if }\;x\ge a
  \end{cases}\;.$
The function $\;G\;$ satisfies all the hypotheses of the previous theorem, so there exists $\lim\limits_{x\to+\infty}g(x)= \lim\limits_{x\to+\infty}G(x)=\cfrac{l}{1-\alpha}\;.$

Corollary 4:
If $\;f:\left[a,+\infty\right[\to\mathbb{R}\;$ is a function on $\left[a,+\infty\right[$ where $a>0$,
$M$ is a positive real number $(M>0)$ such that
$\;\big|f(x)\big|\le Mx\;\;$ for all $\;x\in\left[a,+\infty\right[,$
$\alpha\;$ is a positive real number different from $1\; (\alpha>0,\;\alpha\ne1)$
and $\lim\limits_{x\to+\infty}\cfrac{f(x)-f(\alpha x)}{x}=l\in\mathbb{R}\;,$
then there exists $\lim\limits_{x\to+\infty}\cfrac{f(x)}{x}=\cfrac{l}{1-\alpha}\;.$
Proof:
We consider the function $\;g:\left[a,+\infty\right[\to\mathbb{R}\;$ defined as
$g(x)=\cfrac{f(x)}{x}\quad$ for all $\;x\in\left[a,+\infty\right[\;.$
It results that
$\big|g(x)\big|=\cfrac{\big|f(x)\big|}{x}\le M\quad$ for all $\;x\in\left[a,+\infty\right[\;.$
So the function $\;g(x)\;$ is bounded on $\;\left[a,+\infty\right[\;$ and
$\lim\limits_{x\to+\infty}\big[g(x)-\alpha g(\alpha x)\big]=\lim\limits_{x\to+\infty}\left[\cfrac{f(x)}{x}-\alpha\cfrac{f(\alpha x)}{\alpha x}\right]=$
$=\lim\limits_{x\to+\infty}\cfrac{f(x)-f(\alpha x)}{x}=l\in\mathbb{R}\;.$
Thus, the function $\;g(x)\;$ satisfies all the hypotheses of Corollary 3, hence there exists $\lim\limits_{x\to+\infty}\cfrac{f(x)}{x}= \lim\limits_{x\to+\infty}g(x)=\cfrac{l}{1-\alpha}\;.$
