Using the fact that, $$f'(x)=\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \dfrac{f(x-h)-f(x)}{-h},$$ I was able to prove that if $f \colon (a,b)\to\mathbb{R}$ is differentiable at $x\in (a,b)$, then the limit $$f_s(x)\triangleq\lim_{h\to0}\dfrac{f(x+h)-f(x-h)}{2h},$$ exists. My book then says that the converse does not hold, and I'm given $f(x)=\frac{1}{x^2}$ as a counter-example, with $f_s(0)=0$; however, I get $$\lim_{h\to0}\dfrac{\frac{1}{(x+h)^2}-\frac{1}{(x-h)^2}}{2h}=-\frac{2}{x^3},$$ which is undefined at $x=0$. I would appreciate if someone could tell me what I am misunderstanding. Thank you in advance.
Here is my algebra for the limit:
$\begin{alignat*}{2} \lim_{h\to 0}\dfrac{\frac{1}{(x+h)^2}-\frac{1}{(x-h)^2}}{2h}&=\frac{1}{2}\ \lim_{h\to 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{(x-h)^2}}{h}\\ &=\frac{1}{2}\ \lim_{h\to 0}\frac{-\frac{4hx}{(x-h)^2(h+x)^2}}{\frac{h(h-x)^2}{(x-h)^2}}\\ &=\frac{1}{2}\ \lim_{h\to 0}-\frac{4x}{(h-x)^2(h+x)^2}\\ &=-\left(\frac{1}{2}\right)\frac{4x}{(0-x)^2(0+x)^2}=-\frac{2}{x^3}\\ \end{alignat*}$
For $x=0$ you have: $$f(x+h)=f(0+h)=\frac{1}{h^2}$$ $$f(x+h)=f(0-h)=\frac{1}{h^2}$$ so, $\forall h \neq 0$: $$\frac{f(0+h)-f(0-h)}{2h}=0$$ and in particular: $$f_s(0)=0$$