Consider the following expression $$\lim _{x\to \infty} e^{-\lambda x}\int_0^x e^{\lambda t}f(t)dt$$ where $\lambda>0$ and $f$ is positive, bounded and $\int_0^\infty f(x)dx$ converges. I'd like to characterize this limit, but I'm a little rusty on these things.
First, it should be convergent, because the outer exponent suppresses everything except large values of $t$, where $f$ should be small (for large enough values of $x$).
Moreover, I suspect that the conditions on $f$ are really enough to have that the limit is $0$. If we choose any point $a>0$, then $$l_a \equiv \lim _{x\to \infty} e^{-\lambda x}\int_0^a e^{\lambda t}f(t)dt = 0$$ because then the integral is just a number and the exponent then pushes it to zero. We can now consider a sequence of these $l_a$'s where $a \to \infty$, which will of course have the limit $0$. Then, (here is where I make a logical jump) from Heine's definition of limits, it looks reasonable to think the sequence would have the same limit as the original expression.
I've checked this explicitly with a few examples for $f$, i.e. $f(x) = e^{-x}$ and $f(x) = 1/(1+x^2)$, but I'd like to make this proof formal or otherwise find a counterexample.
We can rewrite your limit to $$ \lim_{x\to \infty} \int_{0}^{x} e^{\lambda(t-x)} f(t)\,dt $$ The $e^{\lambda(t-x)}$ factor is always less than $1$. And for an arbitrary $N>0$ we can even write $$ \begin{align} \int_{0}^{x} e^{\lambda(t-x)} f(t)\,dt &{}\le e^{-\lambda N}\int_{0}^{x-N} f(t)\,dt + \int_{x-N}^x f(t)\,dt \\&{}\le e^{-\lambda N} \int_{0}^{\infty} f(t)\,dt + \int_{x-N}^{\infty} f(t)\,dt \end{align}$$ when $x>N$. The last of these integrals goes to $0$ for $x\to\infty$ because you're assuming $\int_0^\infty f(t)\,dt$ to be finite. So the entire limit must be at most $e^{-\lambda N}\int_0^\infty f(t)\,dt$ -- but since $\lambda N$ can be as large as we want, it is actually $0$.