# Does $\frac{f'(x+h)- \frac{f(x+h)-f(x)}{h}}{h}$ converge uniformly to the second derivative?

I have arrived at an expression

$$\frac{f'(x+h)- \frac{f(x+h)-f(x)}{h}}{h}$$ for a compactly supported function $f \in C_C^{\infty}(\mathbb{R}).$

Now I was asking myself, whether we have uniform convergence to the second derivative of $f$ at point $x.$

Intuitively it is clear that there could be a second derivative involved, as the right summand in the nominator goes to $f'(x)$ and thus we have a difference quotient

$$\frac{f'(x+h)-f'(x)}{h}$$ which goes uniformly to zero, as $|f''|$ is bounded. Unfortunately, in this argument I am taking the limits separately which is not allowed I guess, so is there any way to make this argument rigorous?

Since $f$ is $C^{\infty}$ you have that, using Taylor expansion of $f$ and $f',$

$$f(x+h)=f(x)+f'(x)h+\frac12 f''(x)h^2+\frac16f(x)h^3\mathrm{lot},$$ and

$$f'(x+h)=f'(x)+f''(x)h+\frac12 f'''(x)h^2+\mathrm{lot}.$$ Thus,

$$\frac{f'(x+h)- \frac{f(x+h)-f(x)}{h}}{h}=\frac{f'(x)+f''(x)h-\frac{f'(x)h+\frac12f''(x)h^2}{h}}{h}+\mathrm{lot}$$

$$=\frac{f''(x)h-\frac12f''(x)h}{h}+\mathrm{lot}=\frac12 f''(x)+o(h).$$

It seems, surprisingly, that your expression converges to $f''(x)/2$ not $f''(x)$, as proven by @mfl. For example take $f(x)=x^2$ and compute the limit in $x=0$:

$$\lim_{h \to 0}\frac{2h-\frac{h^2}{h}}{h}=2-1=1$$ but $f''(0)=2$.