Prove that $f'(0)=L$. Let $f$ be continuous at $0$. Suppose $\displaystyle \lim _{x\rightarrow 0} \frac{f(2x)-f(x)}{x} =L$. Prove that $f'(0)=L$.


My Work:

$\displaystyle \Bigg|\frac{f(x)-f(0)}{x}-L\Bigg|=\Bigg|\frac{f(x)-f(2x)+f(2x)-f(0)}{x}-L\Bigg|$

$\displaystyle \leq \Bigg|\frac{f(x)-f(2x)}{x}-L\Bigg|+\Bigg|\frac{f(2x)-f(0)}{x}\Bigg|$. Since $f$ is continuous at $0, |f(2x)-f(0)|$ goes to $0$ as $x$ goes to $0$. Now I want to get rid of $|x|$ of the second term. How can I do it? Can somebody please help me?
Let $\epsilon &gt; 0$ be fixed. Since $\displaystyle\lim_{x \to 0}\dfrac{f(2x)-f(x)}{x} = L$, there exists a $\delta &gt; 0$ such that $(L-\epsilon)y \le f(2y)-f(y) \le (L+\epsilon)y$ for all $0 &lt; |y| &lt; \delta$. 

Since $f(x) - f\left(\dfrac{x}{2^n}\right) = \displaystyle\sum_{k = 1}^{n}f\left(\dfrac{x}{2^{k-1}}\right)-f\left(\dfrac{x}{2^k}\right)$, by applying the above bounds for $y = \dfrac{x}{2^k}$, we have: 

$\displaystyle\sum_{k = 1}^{n}(L-\epsilon)\dfrac{x}{2^k} \le \displaystyle\sum_{k = 1}^{n}f\left(\dfrac{x}{2^{k-1}}\right)-f\left(\dfrac{x}{2^k}\right) \le \displaystyle\sum_{k = 1}^{n}(L+\epsilon)\dfrac{x}{2^k}$

$(L-\epsilon)\left(1-\dfrac{1}{2^n}\right)x \le f(x) - f\left(\dfrac{x}{2^n}\right) \le (L+\epsilon)\left(1-\dfrac{1}{2^n}\right)x$

for all $0 &lt; |x| &lt; \delta$.

Now, what happens if you take the limit as $n \to \infty$?


   In case you didn't see it, I'll work it out. By taking the limit as $n \to \infty$, we get $(L-\epsilon)x \le f(x)-f(0) \le (L+\epsilon)x$ for all $0 &lt; |x| \le \delta$. Hence $f'(0) := \displaystyle\lim_{x \to 0}\dfrac{f(x)-f(0)}{x} = L$, as desired.
We have: $2\dfrac{f(2x)-f(0)}{2x}-\dfrac{f(x)-f(0)}{x} \to 2f'(0) - f'(0) = L\to f'(0) = L$