I have the following function: $$\int^{a+\epsilon}_{a-\epsilon}f(x)\dfrac{1}{\epsilon} dx$$
When I take the limit $\epsilon \to 0$, I want to show that this becomes equivalent to the behavior of the Dirac delta function under the integral: $2\int_{a-\epsilon}^{a+\epsilon}f(x)\delta (x-a)dx=2f(a)$.
I apply the following:
$$\lim_{\epsilon \to 0} \int^{a+\epsilon}_{a-\epsilon}f(x)\dfrac{1}{\epsilon} dx = \lim_{\epsilon \to 0} \dfrac{\int^{a+\epsilon}_{a}f(x)dx}{\epsilon} + \lim_{\epsilon \to 0} \dfrac{\int^{a}_{a-\epsilon}f(x)dx}{\epsilon}=2f(a)$$ by using the Fundamental Theorem of Calculus twice.
Is this derivation correct? Sorry if it looks too trivial, I am little bit exhausted and can't be sure if I am doing it correctly.
As $f$ is continuous,
$$\min_{x \in [a -\epsilon,a]} f(x) \cdot \epsilon \leq \int_{a - \epsilon}^a f(x) \ dx \leq \max_{x \in[a -\epsilon,a]} f(x) \cdot \epsilon$$
Thus $$\min_{x \in [a -\epsilon,a]} f(x) \leq {1 \over \epsilon} \int_{a - \epsilon}^a f(x) \ dx \leq \max_{x \in[a -\epsilon,a]} f(x) $$
Similarly, $$\min_{x \in [a ,a+\epsilon]} f(x) \leq {1 \over \epsilon} \int_{a }^{a+ \epsilon} f(x) \ dx \leq \max_{x \in[a ,a+\epsilon]} f(x) $$
In both cases, as $\epsilon \to 0$, the max and min of $f(x)$ converge to $f(a)$, again by continuity of $f$
Hence
$$\lim_{\epsilon\to 0} \int_{a - \epsilon}^{a+ \epsilon} {f(x)\over\epsilon} \ dx = 2 f(a)$$
We can also rewrite this more simply without the splitting, noting that
$$\min_{x \in [a -\epsilon,a+\epsilon]} f(x) \cdot (2\epsilon) \leq \int_{a - \epsilon}^{a+\epsilon} f(x) \ dx \leq \max_{x \in[a -\epsilon,a+\epsilon]} f(x) \cdot (2\epsilon)$$