sufficient condition for having $f' \to 0$ if $f \to 0$ It is well known the fact that if a real function $ f(t) \to 0$ for $t \to \infty$ then not necessary $f'(t) \to 0$, $f(t) = \frac{\sin(t^2)}{t}$ being a clear counter example. However, I have the following 
$$ f(t)\cdot f'(t) \leq -\alpha \cdot f(t)^2$$ with ($\alpha > 0$, $\alpha \in \mathbb{R}$) some constant. From here it is clear that $f(t) \to 0$ as $t \to \infty$. Is this also enough to conclude that $f'(t) \to 0$? I can not convince my self right now ... However, I think the additional requirement is sufficient:
$$ -\alpha_1 \cdot f(t)^2 \leq f(t)\cdot f'(t) \leq -\alpha_2 \cdot f(t)^2$$ with $\alpha_1 > 0, \alpha_2 > 0$, $\alpha_1, \alpha_2 \in \mathbb{R}$ some constants, since dividing both sides by $f(t)$ already bounds $f'$ between something going to zero ... 
 A: Your condition is equivalent to
$$\frac{1}{2} \frac{d}{dt} \left ( f(t)^{2} \right ) \leq -\alpha f(t)^{2}$$
A substitution $g=f^{2}$ gives $\frac{1}{2} \frac{d}{dt} \left ( g(t) \right ) \leq -\alpha g(t)$ and thus:
$$\frac{d}{dt} ln\left ( g(t) \right ) \leq -2\alpha t$$
$g$ is positive. Use the monoticity of both sides:
$$g(t) \leq C_{1}e^{-2\alpha t}$$
Or
$$f(t) \leq C_{2}e^{-\alpha t}$$
Now if we had your additional requirement then we know $\left | f(t)f'(t) \right | \leq -\alpha \left | f(t)^{2} \right |$ for some suitable $\alpha'$ and then
$$\left | f'(t) \right | \leq -\alpha' C_{2}e^{-\alpha t}$$
So it works.
It is actually sufficient that $\lim_{t\rightarrow \infty }f'(t)$ exists and $f'(t)$ is Riemann integrable. Then $\lim_{t\rightarrow \infty }f'(t)=0$.
Suppose $\forall \epsilon >0, \exists  \delta \in  \mathbb{R}$ such that $t'>\delta \Rightarrow \left | f'(t') - L \right | < \epsilon$. This means, for sufficiently large $\delta$:
$$L-\epsilon<f'(t') <L+\epsilon$$
Integrating from $\delta$ to $t(>\delta)$ with respect to $t'$:
$$(L-\epsilon)(t-\delta) < f(t)-f(\delta) < (L+\epsilon)(t-\delta)$$
The inequality is preserved because it holds for for all $t'\geq \delta$.
Suppose $L$ be nonzero and positive, choose $\delta$ such that the inequality holds for $\epsilon=L/2$. This yields a contradiction as we let $t$ tend to infinity, implying $f(\delta)$ diverges.
Likewise suppose $L$ be nonzero and negative and take $\epsilon = -L/2$ to get a similar result.
Finally, for $L=0$, there is no contradiction and all conditions are satisfied.
