Prove or disprove this assertion Question: If $f$ is differentiable at $x$, then for $\alpha\neq 1$, $f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x+\alpha h)}{h-\alpha h}$.
My attempt: By applying the Secant Method, $f'(x)=\frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}$, can I just replace $x_n$ with $x+h$ and $x_{n-1}$ with $x+\alpha h$?
 A: $$\begin{align}
\lim\limits_{h\to0}\frac{f(x+h)-f(x+\alpha h)}{h-\alpha h}
& = \lim\limits_{h\to0}\frac{f(x+h)-f(x)+f(x)+f(x+\alpha h)}{h-\alpha h}\\
& = \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h-\alpha h}-\lim\limits_{h\to0}\frac{f(x+\alpha h)-f(x)}{h-\alpha h}\\
& = \frac{1}{1-\alpha}\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} - \frac{\alpha}{1-\alpha}\lim\limits_{h\to0}\frac{f(x+\alpha h) - f(x)}{\alpha h}\\
& = \frac{f'(x)}{1-\alpha} - \frac{\alpha f'(x)}{1-\alpha}\\
& = f'(x)\end{align}$$
A: $\require{cancel}$
You can write
\begin{eqnarray}
f(x + \alpha h) &=& f(x + h - (1 - \alpha)h) \\
&=& f(x + h) -(1-\alpha)hf'(x + h) + \frac{1}{2}(1-\alpha)^2h^2f''(x + h) - \frac{1}{6}(1 - \alpha)^3h^3 f'''(x + h) + \cdots
\end{eqnarray}
So that for $\alpha \not= 1$
$$
\frac{f(x + h) - f(x + \alpha h)}{(1 - \alpha)h}  = f'(x + h) + \frac{1}{2}(1 - \alpha) hf''(x + h) - \frac{1}{6}(1 - \alpha)^2h^2 f'''(x + h) + \cdots
$$
In  the limit $h\to 0$ all the terms in the r.h.s vanish, except the first one
\begin{eqnarray}
\lim_{h\to 0}\frac{f(x + h) - f(x + \alpha h)}{(1 - \alpha)h}  &=& \lim_{h\to 0}f'(x + h) + \cancelto{0}{\lim_{h\to 0}\frac{1}{2}(1 - \alpha) hf''(x + h)} \\
&&- \cancelto{0}{\lim_{h\to 0}\frac{1}{6}(1 - \alpha)^2h^2 f'''(x + h)} + \cdots \\
&=& f'(x)
\end{eqnarray}
A: ${{f(x+h)-f(x+\alpha h)}\over{h-\alpha h}}={h\over{h-\alpha h}}{{f(x+h)-f(x)}\over h}-{{\alpha h}\over{h-\alpha h}}{{f(x+\alpha h)-f(x)}\over{\alpha h}}$
$lim_{h\rightarrow 0}{h\over{h-\alpha h}}{{f(x+h)-f(x)}\over h}={1\over{1-\alpha}}f'(x)$
$lim_{h\rightarrow 0}{{\alpha h}\over{h-\alpha h}}{{f(x+\alpha h)-f(x)}\over{\alpha h}}={\alpha\over{1-\alpha}}f'(x)$, so the limit is:
$f'(x)({1\over{1-\alpha}}-{{\alpha}\over{1-\alpha}})=f'(x)$.
