TITLE: Derivative definition
QUESTION [8 upvotes]: Let $f$ be a differentiable function in $x_0$. Calculate the following $\lim$:
$$\lim_{h\to 0}\frac{f(x_0+2h)-f(x_0-h)}{5h}$$
since we know from theory that $f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$, then
I said that $x_0-h=t$ and $x_0+2h=t+3h$ where $3h=k$ so $$\frac{3}{5}\lim_{k\to 0}\frac{f(t+k)-f(t)}{k}=\frac{3}{5}f'(t)=\frac{3}{5}f'(x_0-h)$$

REPLY [5 votes]: $$
\dfrac{f(x_0 + 2h) - f(x_0 - h)}{5h} = \dfrac{2}{5}\dfrac{f(x_0 + 2h) - f(x_0)}{2h} + \frac{1}{5}\dfrac{f(x_0 - h) - f(x_0)}{-h}
$$
Now take the limit as $h \to 0$.