Let $f(t)$ be absolutely continuous, uniformly bounded and such that $\lim_{t\to\infty} f(t)=0$. It should be true that $$\lim_{t\to\infty} \frac{d}{dt} f(t)=0$$ but how can one prove it?
I'm trying to find the formal justification that allows to exchange the limit and the derivative.
The result is actually false. Construct a continuous integrable function $g:\mathbb{R}\rightarrow\lbrack0,\infty)$ such that $$ \liminf_{x\rightarrow\infty}g(x)=0<\limsup_{x\rightarrow\infty}g(x). $$ Note that this is done by taking $g$ to be a piecewise affine function which is zero most of the time and at every $n\in\mathbb{N}$, $g(n)=1$ and the area is the one of a triangle of height one and basis $\frac{1}{n^{2}}$. Then define $$ f(x)=\int_{x}^{\infty}g(t)\,dt. $$ The function $f$ is absolutely continuous since $g$ is integrable, $\lim_{x\rightarrow\infty}f(x)=0$ and $f$ is bounded by $\int_{\mathbb{R}% }g(t)\,dt$ but $f^{\prime}(x)=-g(x)$ does not go to zero.