TITLE: Always a differentiable path through a convergent sequence of points in $\mathbb{R}^n$?
QUESTION [6 upvotes]: I've come up with this question in trying to solve a vaguely related exercise:
If $x_n$ is any sequence of points in $\mathbb{R}^n$ with $x_n \longrightarrow 0$, is there a path $\gamma(t)$, $\gamma(0)=0$ that goes through all $x_n$ and which is differentiable at $t=0$ ?
Thank you.

REPLY [4 votes]: As worded, the answer is yes: take a space-filling path and attach a differentiable piece to it. But this is not what you meant to ask. The question could have been this: Is there a path $\gamma\colon (-1,1)\to\mathbb R^n$ such that (a) $v=\gamma'(0)$ exists and is nonzero; (b) $\gamma(t_n)=x_n$ where $t_n$ is a sequence that decreases to $0$? 
Then the answer is no. Here is an example in two dimensions, with basis vectors $e_1,e_2$: $x_n=\frac{1}{n}e_1$ if $n$ is odd and $x_n=\frac{1}{n}e_2$ if $n$ is even. If $\gamma(t)/t\to v$ as $t\to 0$, then $\frac{\gamma(t)}{|\gamma(t)|}\to \frac{v}{|v|}$. However, $\frac{\gamma(t_n)}{|\gamma(t_n)|}$ alternates between $e_1$ and $e_2$, a contradiction.