$f(x)$ are continues at x=0, "if $\exists \lim \frac{\left\{f(x)-f(-x)\right\}}{x}$ then $f^{\prime}(0)$ exist " are error. how to proof? $f(x)$ are continues at x=0, if $$\exists \lim _{x\to 0} \frac{f(x)-f(-x)}{x}$$ then $f^{\prime}(0)$ exist. how to proof this proposition is error? * * * P.S. As obversely $f(x) = | x |$ are fitting.how to proof it generalization.

If $f$ is even, that means $f(x)=f(-x)$, then $\frac{f(x)-f(-x)}{x}=0$. Hence $\exists \lim _{x\to 0} \frac{f(x)-f(-x)}{x}$.

Suppose that the statement is true. Then we would have:

"each even function is differentiable at $0$".

If $f$ is odd, that means $f(-x)=-f(x),$ then we have $\frac{f(x)-f(-x)}{x}=\frac{2f(x)}{x}$. Since in this case $f(0)=0$ we derive

$\frac{f(x)-f(-x)}{x}=2\frac{f(x)-f(0)}{x-0}$.

Suppose that the statement is true. Then we would have:

"each odd function is differentiable at $0$".