Problem:
I want to understand a proof of the claim given in the title. Suppose we have an initial value problem $\{\dot{x}=f(t,x)~,~x(t_0)=x_0\}$ with continious $f$ and solution $x(t)$.
Proof:
If $M$ is a positive invariant set and $x_0 \in M$, then for $h>0$ small enough $$\mathrm{dist}(x_0+hf(t_0,x_0),M)\leq |x_0+hf(t_0,x_0)-x(t_0+h)|_2.$$ [...]
So for every $\varepsilon > 0$ one can find a $\delta >0$ such that $$\mathrm{dist}(x_0+hf(t_0,x_0),M)\leq \left| h\dot{x}(t_0)-\int_{t_0}^{t_0+h}\dot{x}(s)ds \right| _2\color{red}{\leq}\varepsilon h,$$ whenever $h\in (0,\delta]$.
(So we showed that $\lim_{h \searrow 0}\frac{1}{h}\mathrm{dist}(x+hf(t,x),M)=0$ for every $ t\in\mathbb{R},x\in M$.)
Attempt:
I understand everything except the red part above. I see that
$$\left| h\dot{x}(t_0)-\int_{t_0}^{t_0+h}\dot{x}(s)ds \right| _2\overset{\color{red}{\text{ed.:wrong!}}}{\neq}{\scriptsize h\dot{x}(t_0)-\dot{x}(t_0+h)+\dot{x}(t_0)=\underbrace{(h+1)}_{\to 1}\dot{x}(t_0)-\dot{x}(\underbrace{t_0+h}_{\to t_0})\overset{?}{\to}0}$$
(where $\to$ means $\overset{h\searrow 0}{\longrightarrow}$ to be precise) which might be wrong/unhelpful since there's no $\frac{1}{h}$ in front. On the other hand $\dot{x}$ is continious so there should be no problem when pushing the limit around, so probably that's just not the right approach. I messed up here multiple times, corrected in my answer below.
Questions:
1) I feel like I'm missing out on something completely obvious here, help would be much appreciated. If needed, I can provide the "[...]" above, but I don't feel like that will be important... 2) A few lines below the author says "This [that the subt. cond. holds] is clear for every $x\in \mathring{M}$.". How is that clear?
Got it.
$$\left| h\dot{x}(t_0)-\int_{t_0}^{t_0+h}\dot{x}(s)ds \right| _2=h\dot{x}(t_0)-(x(t_0+h)-x(t_0))=h\bigg(\dot{x}(t_0)-\underbrace{\frac{x(t_0+h)-x(t_0)}{h}}_{\longrightarrow\dot{x}(t_0)\text{ for } h\searrow 0}\bigg)$$
Too much text for such a simple mistake, I feel sorry. Hope noone bothered.