Proof verification: Given $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, show that $\lim _{x\to 0}\frac {f(x)}{x}=0$. Given that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$.
Proof: Let $\epsilon\gt 0$ be fixed.
\begin{align*}
\frac {f(x)}x&=\sum_{k=0}^n\color{blue}{\frac{f(x/2^k)-f(x/2^{k+1})}{x/2^{k+1}}}(1/2^{k+1})+\frac{\color{green}{f(x/2^{n+1})}}{x}\\
&=\sum_{k=0}^n\color{blue}{g_k(x)}(1/2^{k+1})+\frac{\color{green}{h_n(x)}}{x}\\
\text{There exists $N$ such that }\\ \left|\frac {f(x)}x\right|&\le \sum_{k=0}^N|g_k(x)|\frac 1{2^{k+1}}+\epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\frac 1{|x|}\epsilon|x|\\
\text{It follows that } \\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\sum_{k=N+1}^\infty\frac 1{2^{k+1}}+\epsilon\\
\text{Taking $N\to \infty$,it follows that}\\&0\le\liminf \left|\frac {f(x)}x\right|\le \limsup\left|\frac {f(x)}x\right|\le \epsilon\tag 1
\end{align*}
Since $\epsilon\gt 0$ is arbitrary, it follows by $(1)$ that $\liminf \left|\frac {f(x)}x\right|= \limsup\left|\frac {f(x)}x\right|=0=\lim_{x\to 0} \frac{f(x)}x. \;\;\; \blacksquare$
Is my proof correct? Thanks.
 A: We could simplify all this. Let
\begin{equation}{\rho} \left(x\right) = \sup_{\left|y\right|  \leqslant  \left|x\right|} \left|\frac{f \left(2 y\right)-f \left(y\right)}{y}\right|\end{equation}
We have for $n  \geqslant  0$
\begin{equation}\renewcommand{\arraystretch}{1.5}  \begin{array}{rcl}\displaystyle  \left|f \left(x\right)-f \left(\frac{x}{{2}^{n+1}}\right)\right|&=&\displaystyle  \left|\sum _{k = 0}^{n} \left(f \left(\frac{x}{{2}^{k}}\right)-f \left(\frac{x}{{2}^{k+1}}\right)\right)\right|\\
& \leqslant &\displaystyle  \sum _{k = 0}^{n} {\rho} \left(x\right) \frac{\left|x\right|}{{2}^{k+1}}\\
& \leqslant &\displaystyle  \left|x\right| {\rho} \left(x\right) \left(1-\frac{1}{{2}^{n+1}}\right)
\end{array}\end{equation}
By taking the limit of this inequality when $n \rightarrow  \infty $, it follows
that
\begin{equation}\left|\frac{f \left(x\right)}{x}\right|  \leqslant  {\rho} \left(x\right) \mathop{\longrightarrow}\limits_{x \rightarrow  0} 0\end{equation}
A: Nowhere in the question is there an estimate of how small $x$ needs to be to make $\left|\frac{f(x)}x\right|$ small.

Since $\lim\limits_{x\to0}\frac{f(2x)-f(x)}x=0$, for any $\epsilon\gt0$, there is an $x_\epsilon$ so that for $|x|\le x_\epsilon$, we have $\left|\frac{f(2x)-f(x)}{x}\right|\le\epsilon$.
For any $|x|\le x_\epsilon$,
$$
\begin{align}
\left|\frac{f(x)}x\right|
&=\left|\sum_{k=0}^\infty\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}\,2^{-k-1}\right|\tag1\\
&\le\sum_{k=0}^\infty\epsilon\,2^{-k-1}\tag2\\[6pt]
&=\epsilon\tag3
\end{align}
$$
Explanation:
$(1)$: $\sum\limits_{k=0}^\infty\left(f(2^{-k}x)-f(2^{-k-1}x)\right)=f(x)-\lim\limits_{u\to0}f(u)$
$(2)$: $\left|\frac{f(2^{-k}x)-f(2^{-k-1}x)}{2^{-k-1}x}\right|\le\epsilon$
$(3)$: $\sum\limits_{k=0}^\infty2^{-k-1}=1$
A: Let $\epsilon > 0$ be arbitrary, and let $x_\epsilon$ be small enough so that
$
\Bigg| \dfrac{f \big( x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg|
\le \epsilon
$ for all $|x| \le x_\epsilon$.
Then,
$$
\Bigg| \dfrac{f \big(x \big)}{x} \Bigg| =
$$
$$
\dfrac{1}{2} \Bigg|
\dfrac{f \big(x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} +
\dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}}
\Bigg| \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg|
\Bigg] \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {4} \big)}{\frac {x} {4}} \Bigg|
\Bigg]
\Bigg] \le \ldots \le
$$
$$
\dfrac{1}{2} \Bigg[
\epsilon +
\dfrac{1}{2} \Bigg[
\epsilon + \ldots +
\dfrac{1}{2} \Bigg[
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{\frac {x} {2^{n}}} \Bigg|
\Bigg]
\Bigg]
\Bigg] =
$$
$$
\sum_{k=1}^n \dfrac{\epsilon}{2^k} +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \le
$$
$$
\sum_{k=1}^\infty \dfrac{\epsilon}{2^k} +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| =
$$
$$
\epsilon +
\Bigg| \dfrac{f \big(\frac {x} {2^{n}} \big)}{x} \Bigg| \xrightarrow{n \rightarrow \infty}
$$
$$
\epsilon + 0 =
$$
$$
\epsilon
$$
