Is it true that $ \lim_{x\to0}\frac{f'\left(x\right)-\frac{f\left(x\right)-f\left(0\right)}{x}}{x}=\frac{f''\left(0\right)}{2} $

You can use a Taylor expansion: $$ f(x)=f(0)+xf'(0)+x^2f''(0)/2+x^2\sigma(x) $$ where $\lim_{x\to0}\sigma(x)=0$. Then \begin{align} \frac{1}{x}\Bigl(f'(x)-\frac{f(x)-f(0)}{x}\Bigr) &= \frac{1}{x^2}\Bigl(xf'(x)-xf'(0)-x^2f''(0)/2-x^2\sigma(x)\Bigr)\\[6px] &=-\frac{f''(0)}{2}-\sigma(x)+\frac{f'(x)-f'(0)}{x} \end{align} Continuity of the second derivative is not needed; one just needs the (first) derivative exists in a neighborhood of $0$ and is differentiable at $0$.

Tags:

Limits

Calculus

Derivatives

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