Why does $$\lim\limits_{x_2 \to x_0, x_1 \to x_0} \frac{\frac{f(x_2)-f(x_0)}{x_2 - x_0} - \frac{f(x_1)-f(x_0)}{x_1 - x_0}}{x_2 - x_1}$$ result in $\frac{f''(x_0)}2$ instead of $f''(x_0)$? Isn't it the definition of $f''(x_0)$?

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    $\begingroup$ Do you mean $f''$ where you have $f'$? $\endgroup$ – TonyK Mar 4 '16 at 13:25
  • $\begingroup$ @TonyK Oh yes, sorry for that. $\endgroup$ – tartaruga_casco_mole Mar 4 '16 at 14:47
  • $\begingroup$ it isn't the definition of $f''$. In fact, you need that $f''$ is continuous at $x_0$. For a proof see: en.wikipedia.org/wiki/Mean_value_theorem_(divided_differences) $\endgroup$ – user251257 Mar 4 '16 at 14:48
  • $\begingroup$ For the intuition: $\frac{f(x+h)-f(x)}{h}$ is roughly $f'(x+h/2)$ not $f'(x)$ or $f'(x+h)$! $\endgroup$ – user251257 Mar 4 '16 at 15:01

Here's a less rigorous/more intuitive explanation. If we assume $x_1 \le x_0 \le x_2$, you can think of $\frac{f(x_2)-f(x_0)}{x_2-x_0}$ as an approximation for $f'\left( \frac{x_2+x_0}{2} \right)$ and $\frac{f(x_1)-f(x_0)}{x_1-x_0}$ as an approximation for $f'\left( \frac{x_1+x_0}{2} \right)$. The distance between these two derivatives is

$$\frac{x_2+x_0}{2} - \frac{x_1+x_0}{2} = \frac{1}{2}(x_2-x_1)$$

However, you are dividing by twice this amount in your limit. Thus, you will end up with half of $f''(x_0)$.


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