Calculate: $\lim_{x \to 0} \frac{f(3x)+f(5x)-2f(2x)}{x}$

Question

$f$ is a derivable function in $0$, and $f'(0)=a$

Calculate in terms of $a$:

$\lim_{x \to 0} \frac{f(3x)+f(5x)-2f(2x)}{x}$

Answer 1

Hint: \begin{align*} &\dfrac{f(3x)+f(5x)-2f(2x)}{x}\\ &=3\cdot\dfrac{f(3x)-f(0)}{3x}+5\cdot\dfrac{f(5x)-f(0)}{5x}-4\cdot\dfrac{f(2x)-f(0)}{2x}\\ &\rightarrow? \end{align*}

Answer 2

Using a Maclauren series we can approximate $f(x)$ as $$f(x) = f(0)+x\cdot f'(0)+...$$ Since we will view x as small all the higher order terms can be ignored. Plugging this into your equation we get $$\lim_{x\rightarrow 0}\frac{f(0)+ 3x \cdot f'(0)+f(0)+ 5x \cdot f'(0)-2\cdot f(0)- 2\cdot 2x \cdot f'(0)}{x}=\lim_{x\rightarrow 0}\frac{4ax}{x}=4a$$

Answer 3

Let $g(k) = \displaystyle \lim_{x \to 0} \dfrac{f(kx)-f(0)}{kx}\implies L = 3g(3)+5g(5)-2g(2)=3a+5a-2(2a)=4a$