I am stuck with proving a limit which I think should be immediate... I will explain the problem and comment one of my attempts. Let $x: [0, \infty) \to \mathbb{R}^n$ be a differentiable arc with $\lim\limits_{t \to \infty} x(t) = x_0 \in \mathbb{R}^n$, $x(t) \not = x_0, \dot{x}(t) \not = 0, t \geq 0$, of which we know the following limit of tangents exists:
$$ \lim\limits_{t \to \infty} \frac{ \dot{x} (t) }{|| \dot{x} (t) ||} ,$$
say it tends to a certain $L \in S^1$. Now I want to prove $ \lim\limits_{t \to \infty} \frac{ x(t) - x_0}{ || x(t) - x_0 ||}= -L$ (this can be graphically seen to make sense).
My best attempt of a proof was just trying to apply the limit definitions, but I can not get the result... The problem arises when I have:
$$ \left| \frac{x(t)- x_0}{ || x(t)- x_0 ||} - \frac{x(t)- x(s)}{ || x(t)- x(s)||} \right| + \left| \frac{x(t)- x(s)}{ || x(t)- x (s) ||} - \frac{\dot{x} (t)}{ ||\dot{x} (t)||}\right| + \left| \frac{\dot{x} (t)}{ ||\dot{x} (t)||} - L \right|,$$ since the last term is smaller than $\varepsilon /3$ if $t >M_1 >0$, the second term is smaller than $\varepsilon/3$ if $s \in (t- \delta, t + \delta)$, but the first term is smaller than $\varepsilon/3$ if $s > M_2 (t)$, i.e., the constant $M_2$ depends on $t$, so I can not make the three summands "small" at the same time. I have tried to prove that $M_2$ does not depend on $t$ but I have not achieved it, I think it can't be done.
So, any help or suggestion is appreciated, thanks in advance!
Here's one possible approach. Even though the result is quite intuitive, it seems to take a little work to get there! Let $T(t) = \dfrac{\dot x(t)}{\|\dot x(t)\|}$ be the unit tangent vector and $\upsilon(t) = \|\dot x(t)\|$ be the speed at time $t$. Without loss of generality, I'll take $x_0 = 0$.
Note that $$ x(t) = -\int_t^\infty \dot x(t)\,dt = -\int_t^\infty \upsilon(t)T(t)\,dt.$$ We are told that $\lim\limits_{t\to\infty} T(t) = L$ and $\lim\limits_{t\to\infty} x(t) = 0$. Given $0<\varepsilon<1/2$, there is $N$ so that $$\|T(t)-L\|<\varepsilon \quad\text{for } t>N.$$ Now write $T(t) = (T(t)-L) + L$, and so \begin{equation} -x(t) = \int_t^\infty \upsilon(t)(T(t)-L)\,dt + L\int_t^\infty \upsilon(t)\,dt.\tag{$*$} \end{equation} Note that by the triangle inequalities, \begin{align*} \big(\|L\|-\epsilon\big)\int_t^\infty\upsilon(t)\,dt&<\left\|\int_t^\infty \upsilon(t)(T(t)-L)\,dt + L\int_t^\infty \upsilon(t)\,dt\right\|\\ &<\big(\|L\|+\epsilon\big)\int_t^\infty\upsilon(t)\,dt \end{align*} when $t>N$. Since $\|L\|=1$ and substituting ($*$), we obtain \begin{equation} (1-\varepsilon)\int_t^\infty\upsilon(t)\,dt<\|x(t)\| \le \displaystyle\int_t^\infty\upsilon(t)\,dt. \tag{$**$} \end{equation} (Parenthetically, we remark that since $\|x(t)\|\to 0$ as $t\to\infty$, it follows that $\displaystyle\int_t^\infty\upsilon(t)\,dt \to 0$ as $t\to\infty$. This suggests that the curve can't be spiraling around to acquire arclength.)
Now, for $t>N$, it follows from ($*$) that $$\left\|\frac{x(t)}{\int_t^\infty\upsilon(t)\,dt} + L \right\| < \varepsilon.$$ From ($**$), we have \begin{align*} \left\|\frac{x(t)}{\|x(t)\|} + L\right\| &\le \left\|\frac{x(t)}{\|x(t)\|} - \frac{x(t)}{\int_t^\infty\upsilon(t)\,dt}\right\| + \left\|\frac{x(t)}{\int_t^\infty\upsilon(t)\,dt}+L\right\| \\ &< \frac{\varepsilon}{1-\varepsilon}\frac{\|x(t)\|}{\int_t^\infty\upsilon(t)\,dt} + \varepsilon < \frac{\varepsilon}{1-\varepsilon} + \varepsilon < 3\varepsilon. \end{align*} Whew! Since $\varepsilon$ is arbitrary, of course, this proves the desired limit.