Is there any sequence of functions containing functions growing slower than any smooth function? The function 
$f(x)= \begin{cases} 
      e^{-\frac{1}{x}} & x > 0 \\
      0 & x\leq 0 \\
   \end{cases}$
is a smooth function which grows, at $0$, slower than any function of the form $x^n$.
My question is, does there exist a sequence of functions $(f_n)$, infinitely differentiable at $0$ and with $f_n(0)=0$ and $f_n(x)\neq 0$ when $x\neq 0$ for all $n$, with the following property?

For any function $f$ infinitely differentiable at 0 where $f(0)=0$ and $f(x)\neq 0$ when $x\neq 0$ there exists a natural number $N$ such that for all $n\geq N$, we have $\lim_{x\rightarrow 0}\frac{f_n(x)}{f(x)}=0$.

Or is there no such sequence of functions, i.e. for any sequence of functions does there exist a function which grows more slowly than all the functions in the sequence?
 A: There is no such sequence at least if we don't insist on the requirement that $f$ is infinitely often differentiable.
Assume the contrary.
Construct $a_n$ with $\emptyset\neq\left(-a_n,a_n\right)\subseteq f_n^{-1}\left(\left(-\frac1n,\frac1n\right)\right)$ and $a_{n+1}\leq \frac{a_{n}}2$ for $n=1,2,\ldots$. (Possible since the $f_n$ are continuous at $0$.)
Define $f(x) := f_n(x)$ for $x\in (-a_{n},a_{n})\setminus(-a_{n+1},a_{n+1})$.
Now assume $\lim_{x\rightarrow 0}\frac{f_n(x)}{f(x)}=0$ for $n\geq N$. Then we have some $\delta>0$ such that $\left|\frac{f_n(x)}{f(x)}\right|< 0.5$ for all $x\in(-\delta,\delta)$ and $n\geq N$. But we also have some $n\geq N$ with $(-a_{n},a_{n})\subset (-\delta,\delta)$ and $\frac{f_n(x)}{f(x)}=1$ for $x\in(-a_n,a_n)$.
A: There is no such $C^\infty$ function. To prove this, let  $f_n$ satisfy the conditions you specify. WLOG, we can assume all $f_n>0$ on $\mathbb R\setminus \{0\}.$ If that is not true, go to the functions $f_n^2.$ We will construct an even $f\in C^\infty(\mathbb R),$ positive on $\mathbb R\setminus \{0\}$ and strictly increasing on $[0,\infty),$ such that for every $n,$
$$\tag 1\lim_{x\to 0} \frac{f_n(x)}{f(x)}=\infty.$$
For any $g\in C^l(\mathbb R),$ define
$$\|g\|_l = \sum_{j=0}^{l}\sup_{\mathbb R} |D^jg|.$$
If $g_k\in C^\infty(\mathbb R), k = 1,2,\dots,$ and $\sum_{k=1}^{\infty} \|g_k\|_l <\infty$ for each $l,$ then $\sum_{k=1}^{\infty} g_k \in C^\infty(\mathbb R).$ Furthermore, for each $l,$
$$D^l\left(\sum_{k=1}^{\infty} g_k\right ) = \sum_{k=1}^{\infty} D^lg_k.$$
Choose a sequence $a_1>a_2 > \cdots \to 0.$ For each $k,$ set $E_k=\{x:a_{k+1}\le |x|\le a_k\}$ and define $m_k$ to be the smallest of the numbers
$$\min_{E_k} f_1,\min_{E_k} f_2,\dots,\min_{E_k} f_k.$$
Now choose a positive sequence $c_k\to 0$ such that $c_km_k$ strictly decreases to $0.$
For each $k$ we can choose $g_k\in C^\infty(\mathbb R)$ with $g_k>0$ on $(a_{k+1},a_k)$ and $g_k=0$ elsewhere. By multiplying by small enough positive constants we can also require 
$$\|g_k\|_k < 1/2^k \text { and } \int_{a_{k+1}}^{a_k} g_k < c_km_k - c_{k+1}m_{k+1}.$$
We then define
$$g = \sum_{k=1}^{\infty}g_k.$$
This $g\in C^\infty(\mathbb R).$
Finally we get to define $f$ (almost): Define
$$f(x) = \int_0^x g,\,\, x\in \mathbb R.$$
Then $f\in C^\infty(\mathbb R)$ and $f$ is strictly increasing on $[0,\infty).$ Fix $n.$ Claim: If $x\in [a_{k+1},a_k]$ and $k\ge n,$ then
$$\tag 2\frac{f_n(x)}{f(x)} \ge \frac{1}{c_k}.$$
Since $c_k\to 0^+,$ $(2)$ proves $(1),$ at least as $x\to 0^+.$ To prove $(2),$ note $f(x)\le f(a_k).$ Now
$$f(a_k) = \int_0^{a_k} g =\sum_{j=k}^{\infty}\int_{a_{j+1}}^{a_j} g_j < \sum_{j=k}^{\infty}(c_jm_j - c_{j+1}m_{j+1}) = c_km_k.$$
So
$$\frac{f_n(x)}{f(x)}\ge \frac{f_n(x)}{f(a_k)} > \frac{m_k}{c_km_k}=\frac{1}{c_k},$$
proving $(2).$
We're done except for a minor detail. This $f=0$ on $(-\infty,0].$ All we need to do is redefine $f$ for $x\le 0$ by setting $f(x)=f(-x).$ Because all $D^lf(0)=0$ for the original $f,$ the redefined $f\in C^\infty(\mathbb R ).$ Since the $m_k$ took into account the behavior of the $f_n$ to the left of $0,$ we have our counterexample as claimed.
A: There may be some typos in your question. We can simply put, for each $n\in\mathbb{N}$, $f_{n}(x)=1$.
Then, for any $f$ such that $f'(0)$ exists and $f(0)=0$, we automatically have: For each $n$, $\lim_{x\rightarrow0}\frac{f(x)}{f_{n}(x)}=\lim_{x\rightarrow0}f(x)=0$
because $f$ is continuous at $x=0$.
A: You can use 
$$f(x)= \begin{cases} 
      e^{-\frac{1}{x^n}} & x > 0 \\
      0 & x\leq 0 \\
   \end{cases}$$
for $n=1,2,3,4,\ldots$ as a slower and slower growing sequence.
