Generalised limits via derivatives of integrals?

Question

Assuming that $f$ is a continuous function, we have that

$$f(x) = \frac{d}{dx}\int f(t)\,dt.$$

Assuming instead that $f$ has a removable singularity at $x=a$, and is otherwise continuous, we have that

$$\lim_{x\to a}f(x) = \left.\frac{d}{dx} \int f(t)\,dt\right|_{x=a},$$ where the integral is either an improper Riemann integral, or a Lebesgue or Gauge integral.

But what if the singularity at $x=a$ is not removable? For instance, we have that

$$\left.\frac{d}{dx}\int \sin\Big(\frac{1}{t}\Big)\,dt\right|_{x=a}=0,$$

but the singularity at $x=0$ of $\sin(1/x)$ is not removable.

How does this relate to other notions of generalised limit, such as Cesaro's or Abel's?

[edit]

For clarification, the procedure above is, given a function $f$, and a real number $a$, to consider the following limit $$\lim_{\epsilon \to 0} \frac{\int_a^{a+\epsilon}f(t)\,dt}{\epsilon}.$$ The result you get from $f(x) = \sin(1/x)$ and $a = 0$ comes from the fact that the integral of $f$ has a removable singularity at $x=0$, on which $\int f$ can still be differentiated, albeit the resulting derivative is discontinuous.

Answer 1

I think that the concept of the limit of a distribution might be relevant. This can be defined elementarily as follows: a distribution $f$ on, say, $]0,1]$ has limit $a$ at zero if it has the form $D^p F$ (distributional derivative) near zero, where $$\lim_{x\to0}\frac{F(x)}{x^p}=\frac a{p!}.$$

Every continuous on ${]0,1}$ function is a distribution and it can happen that it has a limit in the distributional sense, but not in the classical one. Thus we have $\lim_{x\to \infty} \sin x=0$ in the distributional sense (forgive me for switching to limits at infinity for the purpose of this example).

The case where $p=1$ gives that the limit of a continuous $f$ is $a$ if $\lim_{x\to 0} \frac {F(x)}x = 0$ where $F$is a primitive (application: $f(x)= \sin{\frac 1 x}$).