The fractional order derivative approach above We recall the Riemann-Liouville left sided fractional $s$-order derivative is defined as, in interval $(0,1)$, where $0<s<1$,
$$
D^s_{[0,x]}f(x):=\frac{1}{\Gamma(1-s)}\frac d{dx}\int_0^x \frac{f(y)}{(x-y)^s}dy\tag 1
$$
for $x\in(0,1)$, where $\Gamma$ denotes the Gamma function.
It is clear that for $s=0$, $(1)$ reduces to $f(x)$. I am wondering what if $s=1$? should it give $f'(x)$? I suppose so, at least from the numerical point of view. But I found it hard to prove analyliticly... More precisely, what would $(1)$ become if we send $s\nearrow 1$? converge to $f'(x)$? If yes, in what capacity?
To be precise. For a function $f\in C_c^\infty(0,1)$, do we have 
$$
\lim_{s\nearrow 1} \int_0^1|D^s_{[0,x]}f(x)-f'(x)|dx =0?
$$
Any helps?
 A: Actually, it doesn't exist for $s=1$.  Indeed, this is why we have $0<s<1$.  You will notice that we have will have $\Gamma(0)$.  Indeed, $\Gamma(0)$ doesn't exist, and as $s\to1$, $\frac1{\Gamma(1-s)}\to0$.  Just the same, you will notice that if $f(x)\ne0$, then $\int_0^x\frac{f(y)}{(x-y)^s}dy\to\pm\infty$.  Thus we result with the following indeterminate form:
$$\frac1{\Gamma(1-s)}\frac d{dx}\int_0^x\frac{f(y)}{(x-y)^s}dy\stackrel{s\to1}=0\times\infty$$
which is quite unfortunate.  However, it is noticeable that if we were to evaluate the limit, we'd end up with the following:
$$\lim_{s\to1^-}\frac1{\Gamma(1-s)}\frac d{dx}\int_0^x\frac{f(y)}{(x-y)^s}dy=f'(x)$$
just as you would expect.
Now, if $u,v$ were functions of $x$, one might apply the multivariate chain rule as follows:
$$\frac d{dx}\int_0^uf(y)(v-y)^sdy=u'f(u)(v-u)^s+sv'\int_0^uf(y)(v-y)^{s-1}dy$$
whenever $0<s<1$, $u(x)=v(x)=x$, we end up with
$$\frac d{dx}\int_0^xf(y)(x-y)^sdy=s\int_0^xf(y)(x-y)^{s-1}dy$$
This result means that
$$\frac1{\Gamma(1-s)}\frac d{dx}\int_0^xf(y)(x-y)^{-s}dy=\frac1{\Gamma(2-s)}\frac{d^2}{dx^2}\int_0^xf(y)(x-y)^{1-s}dy$$
Now letting $s\to1$ on the RHS gives us

$$\lim_{s\to1}\frac1{\Gamma(2-s)}\frac{d^2}{dx^2}\int_0^xf(y)(x-y)^{1-s}dy=\frac1{\Gamma(1)}\frac{d^2}{dx^2}\int_0^xf(y)dy=f'(x)$$

You may also want to check the Wikipedia or Google for other references.
