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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 $x+\alpha h$?

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$$\begin{align} \lim\limits_{h\to0}\frac{f(x+h)-f(x+\alpha h)}{h-\alpha h} & = \lim\limits_{h\to0}\frac{f(x+h)-f(x)+f(x)+f(x+\alpha h)}{h-\alpha h}\\ & = \lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h-\alpha h}-\lim\limits_{h\to0}\frac{f(x+\alpha h)-f(x)}{h-\alpha h}\\ & = \frac{1}{1-\alpha}\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h} - \frac{\alpha}{1-\alpha}\lim\limits_{h\to0}\frac{f(x+\alpha h) - f(x)}{\alpha h}\\ & = \frac{f'(x)}{1-\alpha} - \frac{\alpha f'(x)}{1-\alpha}\\ & = f'(x)\end{align}$$

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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}

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${{f(x+h)-f(x+\alpha h)}\over{h-\alpha h}}={h\over{h-\alpha h}}{{f(x+h)-f(x)}\over h}-{{\alpha h}\over{h-\alpha h}}{{f(x+\alpha h)-f(x)}\over{\alpha h}}$

$lim_{h\rightarrow 0}{h\over{h-\alpha h}}{{f(x+h)-f(x)}\over h}={1\over{1-\alpha}}f'(x)$

$lim_{h\rightarrow 0}{{\alpha h}\over{h-\alpha h}}{{f(x+\alpha h)-f(x)}\over{\alpha h}}={\alpha\over{1-\alpha}}f'(x)$, so the limit is:

$f'(x)({1\over{1-\alpha}}-{{\alpha}\over{1-\alpha}})=f'(x)$.

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