What's the fallacy here?

Question

Can anybody find the fallacy in the below argument? I have used the symmetric definition of the double derivative and that gives me a crazy consequence...

$$f''(x_0)=\lim_{x\to x_0}\frac{f(x)-2f(x_0)+f(x_0-(x-x_0))}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{f(x)-f(x_0)+f(x_0-(x-x_0))-f(x_0)}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{f(x)-f(x_0)+f(2x_0-x)-f(x_0)}{(x-x_0)^2}$$ Now let $g(x)=f(2x_0-x)$, then $g'(x_0)=-f'(x_0)$ $$f''(x)=\lim_{x\to x_0}\frac{f(x)-f(x_0)+g(x)-g(x_0)}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{f(x)-f(x_0)+(x-x_0)g'(x_0)}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{f(x)-f(x_0)-(x-x_0)f'(x_0)}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{\sum_0^\infty\frac{f^{(i)}(x_0)}{i!}(x-x_0)^i-f(x_0)+(x-x_0)g'(x_0)}{(x-x_0)^2}$$ $$=\lim_{x\to x_0}\frac{\sum_2^\infty\frac{f^{(i)}(x_0)}{i!}(x-x_0)^i}{(x-x_0)^2}$$ $$=\frac{1}{2} f''(x)$$

Answer 1

The error is in the step where you replace $g(x)-g(x_0)$ by its local linear approximation, $(x-x_0)g'(x_0)$. You need to keep terms up to second order: $$ g(x)-g(x_0)\approx(x-x_0)g'(x_0)+\frac{1}{2}(x-x_0)^2g''(x_0). $$ In other words, you cannot assume that this second order term vanishes in the limit.

Answer 2

Adding to TheSimpliFire's note, I suspect you have confused yourself by using $x_0$ notation. Your definition of $f''$ treats the real variable as $x_0$, but later you subtly to switch to $x$.

After your definition of $f''$, you should have defined $g$ as $g(x_0) = f(2x_0 - x)$, not $g(x)$, and you won't get the contradiction.

In effect, you have defined a totally different curve $g$, and stated a point of intersection, and then falsely assumed that the local derivatives of the two curves $f$ and $g$ have to align.