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Two cars $X$ and $Y$ are in a race. The position of car $X$ as a function of time is described by $x(t)$. The position of car $Y$ is described by $y(t)$. Initially $x(t),y(t)>0$. Whenever $x(t)=y(t)$,we have $x'(t)≤y'(t)$. Further we know that $y'(t) < 0$ for all times $t ≥ 0$. Prove/ Disprove: Car $X$ either crosses the position $0$ in finite time, or $\lim_{t→∞} x(t) = 0$. Assume that $x(t)$ and $y(t)$ are infinitely differentiable.

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    It seems that something you've written is wrong. First, you say that "initially $x(t),y(t)>0$" then you say $y(t) < 0 \forall t \geq 0$. What time is "initially" then...before $t=0$? Also, the statement "whenever $x(t)=y(t)$, we have $x(t) \leq y(t)$" is uninformative -- that's always true. Lastly, let $y(t)$ be any decreasing function and $x(t)$ be any increasing function, clearly it doesn't cross $0$ and it's limit won't be $0$ either. So the proof is trivial. Next, using a Markov chain to discuss deterministic functions doesn't make much sense either. – 2017-02-23
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    its derivative of y(t) < 0. I'm extremely sorry if that dot is not visible. – 2017-02-23

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Perhaps I am misunderstanding the problem, but consider the counterexample to disprove the statement.

$$y(t) = 2-\exp(t)$$ $$ x(t) = 3 + \exp(t).$$

Then we have of course that $$y(0),x(0)>0.$$ Since $$x(t)\neq y(t), \quad t\in \mathbb{R},$$ we don't have to worry about the condition ordering the derivatives. And $$y'(t) = -\exp(t) < 0, \qquad \forall t\geq 0.$$

Clearly though, $x(t) > 0, \; \forall t$ and so it never crosses the $0$, additionally, $$\lim_{t\to \infty}x(t) = \infty \neq 0.$$

Are you missing other conditions in the problem statement? Otherwise, it seems to be false.

Also, everything is deterministic, so it is still unclear why you are trying to use Markov chains with this problem.