"if $f$ is integrable on $[0,A]$ for every $A>0$, and $\lim_{x\to\infty}f(x)=1$, then $$ \lim_{t\to 0^+} t\int_{0}^{\infty} e^{-tx}f(x)\: dx $$ exists"
(I'm convinced it's true, some examples suggest that the answer is 1 for every function with these properties, but how to prove it?)
For each $\epsilon > 0$, break up the integral at $A$ such that $$1-\epsilon \leq f(x) \leq 1+ \epsilon \quad \text { for }x\geq A.$$
The first part of the integral $$\lim_{t\rightarrow 0^+} \int_0^A t e^{-tx} f(x) dx = 0 $$ from Lebesgue dominated convergence theorem with bounding function $|f|$ when $t<1$.
On the 2nd part $$(1-\epsilon) \int_A^\infty te^{-tx} dx \leq\int_A^\infty te^{-tx} f(x) dx\leq (1+\epsilon)\int_A^\infty te^{-tx} dx \quad\quad (1)$$ and using integration from calculus $$\int_A^\infty te^{-tx} dx = -e^{-tx} \bigg|_A^\infty = e^{-tA}.$$ Take the limit as $t$ goes to $0^+$ in $(1)$, we have $$(1-\epsilon) \lim_{t\rightarrow 0^+} e^{-tA} \leq \lim_{t\rightarrow 0^+}\int_A^\infty te^{-tx} f(x) dx\leq (1+\epsilon) \lim_{t\rightarrow 0^+} e^{-tA}$$
$$1-\epsilon \leq \lim_{t\rightarrow 0^+}\int_A^\infty te^{-tx} f(x) dx\leq 1+\epsilon.$$