Let $\lambda, h : \mathbb{R} \to \mathbb{R}$ and suppose $\int_{-\infty}^\infty h(t)\,dt$ exists and $\lim_{t \to \infty}\lambda(t)$ exists. I want to show that the limit of their convolution equals the product of the integral under one and the limit of the other:
$$\lim_{t \to \infty} (\lambda*h)(t) = \lim_{t \to \infty} \int_{-\infty}^\infty h(\tau) \lambda(t-\tau) \,d\tau = \left(\int_{-\infty}^\infty h(t)\,dt\right)\left( \lim_{t \to \infty} \lambda(t) \right). $$ The reason why I think this should be the case is that since $\int_{-\infty}^\infty h(t)\,dt$ is finite, $h(t)$ has to converge to 0 as $t \to \infty$ and as $t \to -\infty$, thus the only parts of $\lambda$ that contribute to the convolution integral are those "finitely near" to $t$, which as $t \to \infty$, become a constant value, $\lim_{t \to \infty}\lambda(t)$. So in the limit this constant can be put in front of the integral and the result follows. However I don't know how I would go about formally proving this (and even whether it is always true given the conditions I gave).
The reason why I want to show this is because in my case, both $\lambda$ and $h$ map from positive reals to positive reals, and $\lambda$ is the "input" and $h$ is the "impulse response", and I basically want to have a way of showing what the limiting/steady-state response will be if I know what the input will converge to be.
