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?
 A: Let $\epsilon > 0$ be fixed. Since $\displaystyle\lim_{x \to 0}\dfrac{f(2x)-f(x)}{x} = L$, there exists a $\delta > 0$ such that $(L-\epsilon)y \le f(2y)-f(y) \le (L+\epsilon)y$ for all $0 < |y| < \delta$. 
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 $0 < |x| < \delta$.
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 $0 < |x| \le \delta$. Hence $f'(0) := \displaystyle\lim_{x \to 0}\dfrac{f(x)-f(0)}{x} = L$, as desired.

A: Take any $\varepsilon >0$ and let $\delta >0$ be a number such that $$\left| \frac{f(2t) -f(t)}{t} -L\right|\leq \varepsilon $$ for all $|t|\leq \delta .$ Take any $t$ with $0<|t|\leq \delta $ then we have $$\left|\frac{f(t) -f(0)}{t} -L\right | =\left|\frac{\sum_{k=0}^n \left(f\left(\frac{t}{2^k} \right) - f\left(\frac{t}{2^{k+1}} \right) \right)}{t}+\frac{f\left(\frac{t}{2^{k+1}}\right) -f(0)}{t} -L\right|\leq \sum_{k=0}^n \frac{1}{2^{k+1}}\left|\frac{f\left(\frac{t}{2^k} \right) - f\left(\frac{t}{2^{k+1}} \right) }{\frac{t}{2^{k+1}}}-L\right|+\frac{|L|}{2^{n+1}} +\left|\frac{f\left(\frac{t}{2^{k+1}}\right)- f(0)}{t}\right|.$$ Now take $n$ such a big that $2^{-(n+1)}|L|\leq \varepsilon $ and $\left|f\left(\frac{t}{2^{k+1}}\right)- f(0)\right|\leq \varepsilon |t|$ then $$\left|\frac{f(t) -f(0)}{t} -L\right |\leq \sum_{k=0}^{n} \frac{1}{2^{k+1}}\varepsilon +\varepsilon +\varepsilon <3\varepsilon$$ therefore $f'(0)$ exists.
