A subsequence converging to an analytic curve

This is a question that confronted me while I was trying to find a proof of a different result. Consider a sequence $x_n \in \mathbb{R}^m$ with $x_n \to 0$. Under what circumstances does a subsequence converge to an analytic curve $\gamma:[0,\varepsilon) \to \mathbb{R}^m$ with $\gamma(0)=0$? Let me make this notion precise: the sequence must converge according to all derivatives, that is, faster than any power of $r = \|x\|$. It should be noted that the eventual subsequence we extract need not lie on the curve, because of the existence of nonanalytic functions which converge more quickly than any polynomial. By the compactness of $S^{m-1}$ we can pass to a subsequence such that $s_1^n = \|x_n\|$ is converging to some point $s_1 \in S^{m-1}$. So we begin to construct a curve by starting with \[ \gamma(t) = tv_1, \] where $v_1=s_1$. Next, let $s^2_n$ be the intersection of $\gamma(t) = v_1t +v_2^nt^2$ with $S^{n-1}$, where the vector $v_2^n$ is chosen such that $\gamma(t)$ contains $x_n$. Passing to a subsequence, we have $s^2_n \to s_2$. We then have the curve \[ \gamma(t) = v_1t +v_2t^2, \] where $v_2$ is chosen so that the curve intersects $S^{m-1}$ at $s_2$. We assume for the moment that such a $v_2$ exists, and examine this assumption shortly. Note that the distance from $x_n$ to the curve is bounded by $ct_n^2$, where $\gamma(t_n)$ is the point of the curve closest to $x_n$, and since $s^2_n \to s_2$, $c$ can be made as small as we like by truncating the sequence. Iterating this process, we arrive at a curve \[ \gamma(t) = v_1t +v_2t^2 + \ldots + v_kt^k, \] and a subsequence $x_n$ such that $d(x_n,\gamma(t_n)) \leq ct_n^k \leq cr_n^k$, where $r_n =||x_n||$. Now we return to the question of whether the vector $v_2$ (and $v_3,\ldots,v_k$) actually exists. It can happen that as $s^2_n \to s_2$, $v_2^n$ becomes unbounded and consequently $v_2$ doesn't exist. This means that the sequence is converging to $v_1$ slower than $t^2$. This corresponds to the case where the curve must be written as a Puiseux series, rather than a Taylor series. In this case, we multiply the Taylor series by $t$, i.e. we consider \[ \gamma(t) = v_1t^2 +v_2t^3 + \ldots + v_kt^{k+1}, \] and again try to construct $v_2$. It could happen that $v_2 = 0$, in which case we attempt to construct $v_3$, continuing to multiply by $t$ whenever a vector fails to exist. Eventually we will obtain the first two non-zero terms of our curve \[ \gamma(t) = v_1t^{1+l} + v_jt^{j+l} \] for some $j \geq 2$. Otherwise, all subsequences of the sequence $x_n$ must be asymptoting to $v_1$ slower than any rational power $\rho$ of $r$, $\rho > 1$, $\rho \to 1$. From here, we can construct all remaining terms using our original process, i.e. all remaining vectors $v_i$ will exist. Thus, after multiplying by $t$ enough times, we will eventually be able to construct a curve \[ \gamma(t) = v_1t^{1+l} +v_2t^{2+l} + \ldots + v_kt^{k}, \] which satisfies our requirements. However, it's unfortunately possible for $x_n$ to be asymptoting to $v_1$ slower than any rational power or $r$. But, it means it's converging to zero so fast relative to its rate of rotation that I believe it is possible to find a higher dimensional manifold that it is asymptoting to. I'm working on making this notion precise in a paper I'm working on.