Does comparing a function against $x^N$ ensure $N$ times differentiable? If $$\lim_{x \to 0} \frac{f(x)}{x^N} = 0,$$
does that automatically ensure that $f$ has an $N^\text{th}$ derivative at $0$? Noting that that would require an $(N-1)^\text{st}$ derivative in an interval around $0$, it seems unlikely to me that this implication is true, but I also can't find a counterexample.
Note that the corresponding question for a more general Taylor polynomial would be: does
$$\lim_{\Delta x \to 0} \frac{f(x_0+\Delta x)-g(x_0 + \Delta x)}{(\Delta x)^N} = 0,$$
with $g$ a degree $N$ polynomial, force $f$ to be $N$ times differentiable at $x_0$ with Taylor polynomial $g(x)$? This question is equivalent by an appropriate substitution.
Note that the statement is true for $N=1$, and can be considered true for $N=0$ if you define $0$ times differentiable at a point as continuous at that point. So a counterexample would have to be constructed for $N \ge 2$.
 A: Your question is related to a concept known as the Peano derivative. 
Definition. For an open set $U\subset\mathbb{R}$ we say $f:U\to\mathbb{R}$ is $n$ times Peano differentiable at $a\in U$ if there exist $f_m(a)\in\mathbb{R}$ for all $m\leq n$ such that
$$\lim_{x\to a}\frac{f(x)-\sum_{m\leq n}\frac{f_m(a)}{m!}(x-a)^m}{(x-a)^n}=0$$ 
Example. A classic example is $f(x)=x^{n+1}\sin(x^{-n})$ for $n\in\mathbb{N}$ and $f(0)=0$. 
It is an easy exercise that $f$ is $n$ times Peano differentiable at zero, by picking $f_m(0)=0$ for all $m\leq n$. For in that case,
$$\lim_{x\to0}\frac{x^{n+1}\sin(x^{-n})}{x^n}=\lim_{x\to0}x\sin(x^{-n})=0$$
But is $f$ differentiable up to degree $n$ at zero? Note the first derivative is $$f'(x)=(n+1)x^n\sin(x^{-n})-n\cos(x^{-n})$$ for $x\neq0$ and $f'(0)=0$. But this function is not even continuous at zero, so $f$ has no derivative of degree two or higher there.
History. The concept does date back to Peano, though many have forgotten it. See section 2.5, "Peano, de La Valleé Poussin and Generalized Derivatives" by Jean Mawhin in Giuseppe Peano between Mathematics and Logic, edited by Fulvia Skof.
A: This is a bizarre way of doing it, so I'm hoping someone has a more instructive solution.
Define $f$ to be an even function on $(-1, 1)$ with $f(x) = x^n$ on $\left[\frac{1}{2^{n+1}}, \frac{1}{2^n}\right)$ and $f(0) = 0$.
Then, for any $N$, we have $\lim_{x\rightarrow 0} \frac{f(x)}{x^N} = 0$, but $f'$ can't exist on a neighborhood of zero, since $f$ is discontinuous on $\{\frac{1}{2^n}: n \in \mathbb{N}\}$.
A: The problem can be solved by going to the basics.
The given limit condition is essentially an information about the behavior of the function $f$ in neighborhood of $0$ and thus is a local information. It can not be used to infer anything about the local behavior of function at some other point.
If we are also given that $f(0)=0$ then we can infer the continuity and differentiability of $f$ at $0$. But one can't in general infer anything about $f$ at other points (not even continuity let alone differentiability). The question of higher derivatives at $0$ does not arise because it requires the existence of derivatives in a neighborhood of $0$ and not at just $0$.
While studying analysis/calculus one should not in general assume more than what is available as standard result and things which can be proved using them. Unfortunately calculus is one subject where people assume many things without thinking through. An example: If derivative is positive on some interval then function is strictly increasing on that interval (True); people also assume the converse: If a differentiable function is strictly increasing on an interval then derivative is positive on that interval (False). 
