Prove that $f'(0)$ exists and $f'(0) = b/(a - 1)$ Problem:
  If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.


This approach is definitely wrong:  

\begin{align}
b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\
&amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\
&amp;=af'(0)-f'(0)\\
&amp;=(a-1)f'(0)
\end{align}

I will show you a case why this approach is wrong:


  \[f(x)= \begin{cases}
1,&amp;x\neq0\\
0,&amp;x=0
\end{cases}\]
  $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$
  but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$


Does anyone know how to prove it? Thanks in advance!
This is a tricky question and the solution is somewhat non-obvious. We know that $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ and hence $$f(ax) - f(x) = bx + xg(x)$$ where $g(x) \to 0$ as $x \to 0$. Replacing $x$ by $x/a$ we get $$f(x) - f(x/a) = bx/a + (x/a)g(x/a)$$ Replacing $x$ by $x/a^{k - 1}$ we get $$f(x/a^{k - 1}) - f(x/a^{k}) = bx/a^{k} + (x/a^{k})g(x/a^{k})$$ Adding such equations for $k = 1, 2, \ldots, n$ we get $$f(x) - f(x/a^{n}) = bx\sum_{k = 1}^{n}\frac{1}{a^{k}} + x\sum_{k = 1}^{n}\frac{g(x/a^{k})}{a^{k}}$$ Letting $n \to \infty$ and using sum of infinite GP (remember it converges because $|a| &gt; 1$) and noting that $f$ is continuous at $x = 0$, we get $$f(x) - f(0) = \frac{bx}{a - 1} + x\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ Dividing by $x$ and letting $x \to 0$ we get $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \frac{b}{a - 1} + \lim_{x \to 0}\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$

The sum $$\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}$$ tends to $0$ as $x \to 0$ because $g(x) \to 0$. The proof is not difficult but perhaps not too obvious. Here is one way to do it. Since $g(x)\to 0$ as $x \to 0$, it follows that for any $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that $|g(x)| &lt; \epsilon$ for all $x$ with $0 &lt;|x| &lt; \delta$. Since $|a| &gt; 1$ it follows that $|x/a^{k}| &lt; \delta$ if $|x| &lt; \delta$ and therefore $|g(x/a^{k})| &lt; \epsilon$. Thus if $0 &lt; |x| &lt; \delta$ we have $$\left|\sum_{k = 1}^{\infty}\frac{g(x/a^{k})}{a^{k}}\right| &lt; \sum_{k = 1}^{\infty}\frac{\epsilon}{|a|^{k}} = \frac{\epsilon}{|a| - 1}$$ and thus the sum tends to $0$ as $x \to 0$.

Hence $f'(0) = b/(a - 1)$.



BTW the result in question holds even if $0 &lt; |a| &lt; 1$. Let $c = 1/a$ so that $|c| &gt; 1$. Now we have $$\lim_{x \to 0}\frac{f(ax) - f(x)}{x} = b$$ implies that $$\lim_{t \to 0}\frac{f(ct) - f(t)}{t} = -bc$$ (just put $ax = t$). Hence by what we have proved above it follows that $$f'(0) = \frac{-bc}{c - 1} = \frac{b}{a - 1}$$ Note that if $a = 1$ then $b = 0$ trivially and we can't say anything about $f'(0)$. And if $a = -1$ then $f(x) = |x|$ provides a counter-example. If $a = 0$ then the result holds trivially by definition of derivative. Hence the result in question holds if and only if $|a| \neq 1$.
Hint: You're very close. 

Write the expression as $$a\frac{f(ax)-f(0)}{ax}-\frac{f(x)-f(0)}{x}$$ Note that $x\to 0$ if and only if $ax\to 0$ (since $a\neq 0$).

Can you see it from this?