Taking the limit of the bounds of an integral I need to prove that
$$\lim_{{\Delta \omega}\rightarrow 0^+} \int_{-\pi/\Delta \omega}^{\pi/\Delta \omega} f(\Delta \omega, t) dt = \lim_{{\Delta \omega}\rightarrow 0^+} \int_{-\infty}^{\infty} f(\Delta \omega, t) dt\label{eq:1}$$
Now, I know the composite limit theorem, which states that:
$$\lim_{u\rightarrow M} f(u) = \lim_{x\rightarrow a} f(g(x)) $$ Given that the following holds:
$$\lim_{x\rightarrow a} g(x) = M , \hspace{5mm} \lim_{u\rightarrow M} f(u) = f(M).$$
One can also derive the following property:
$$ \lim_{x\rightarrow a} \left[ f(x) g(x) \right] = \lim_{x\rightarrow a} \left[ K g(x) \right]  $$
Given that $$\lim_{x\rightarrow a}f(x) = K, \hspace{5mm} \lim_{x\rightarrow a}f(x) = L.$$
Is it possible to prove the first equality using these two limit theorems? I'm having doubts about applying the composite limit theorem, as $\int_{-\pi/\Delta \omega}^{\pi/\Delta \omega} f(\Delta \omega, t) dt$ is a functional and not a function. Does there exist an equivalent composite limit theorem for functionals?
 A: I think you should impose some conditions on $f$.
For example, define $f : \mathbb R^2 \to \mathbb R$ by $f(\Delta w, t)= \Delta w\exp\left(-(\Delta w)^2 t^2 \right)   $.
Then for $\Delta w >0 $\begin{align} \int_{-\pi/\Delta w}^{\pi/\Delta w}f(\Delta w, t) dt &= \Delta w\int_{-\pi/\Delta w}^{\pi/\Delta w}\exp \left( -(\Delta w)^2 t^2 \right)dt \\ &= \Delta w\int_{-\pi}^{\pi}\exp\left(-s^2\right)\frac{1}{\Delta w}ds \\&=\int_{-\pi}^{\pi}\exp\left(-s^2\right)ds\end{align}
Here change of variables $s=(\Delta w)t$ is used.
On the other hand, \begin{align} \int_{-\infty}^{\infty}f(\Delta w, t) dt &= \Delta w\int_{-\infty}^{\infty}\exp \left( -(\Delta w)^2 t^2 \right)dt \\ &= \Delta w\int_{-\infty}^{\infty}\exp\left(-s^2\right)\frac{1}{\Delta w}ds \\&=\int_{-\infty}^{\infty}\exp\left(-s^2\right)ds\end{align}
Both integration are independent of $\Delta w$. Therefore $$\lim_{\Delta w \to 0+}\int_{-\pi/\Delta w}^{\pi/\Delta w}f(\Delta w, t) dt = \int_{-\pi}^{\pi}\exp\left(-s^2\right)ds \neq \int_{-\infty}^{\infty}\exp\left(-s^2\right)ds = \lim_{\Delta w \to 0+} \int_{-\infty}^{\infty}f(\Delta w, t) dt $$
