Prove that $f[x_0,x_0,x_1]=f'[x_0,x_1]$ for divided difference

Question

Prove that $f[x_0,x_0,x_1]=f'[x_0,x_1]$ for divided difference

Attempt

RHS \begin{align*} f'[x_0,x_1]&=\lim_{\epsilon\rightarrow 0}\frac{f[x_0+\epsilon,z_1]-f[x_0,x_1]}{\epsilon} \end{align*}

LHS \begin{align*} f[x_0,x_0,x_1]&=\lim_{\epsilon\rightarrow 0}f[x_0,x_0+\epsilon,x_1]\\ &=\lim_{\epsilon\rightarrow 0}\frac{f[x_0,x_0+\epsilon]-f[x_0+\epsilon,x_1]}{x_0-x_1}\\ &=\frac{\frac{f(x_0+\epsilon)-f(x_0)}{\epsilon}-\frac{f(x_0+\epsilon)-f(x_1)}{x_0+\epsilon-x_1}}{x_0-x_1} \end{align*}

How to show that they are equal?

Answer 1

Note that $$ \frac{f[x_0+\epsilon,x_1]-f[x_0,x_1]}{\epsilon}=f[x_0+\epsilon,x_1,x_0] $$ and that the order of the argument points does not matter in divided differences.