How well-behaved are $C^{\infty}$ functions? 
Let $u \in C^{\infty}(\mathbb{R})$ and suppose $u(x_0) = 0$. Does there always exist some $\epsilon > 0$ such that one of the following three conditions is satisfied?
  
  
*
  
*$f(x) > 0$ for every $x \in (x_0, x_0 + \epsilon)$
  
*$f(x) < 0$ for every $x \in (x_0, x_0 + \epsilon)$
  
*$f(x) = 0$ for every $x \in (x_0, x_0 + \epsilon)$

My thoughts
I know that this isn't true if we just require differentiability, thanks to the famous example $x^2 \cdot \sin(1/x)$. In fact, simply choosing $n$ large enough puts $x^n \cdot \sin(1/x)$ into any class $C^k(\mathbb{R})$ for a fixed finite $k$.
However, I'm not sure if requiring $C^{\infty}$ may circumvent this type of scenario. Perhaps an infinite smoothness will get rid of any such kinks? The example I gave seems to suggest that we'd need to have "$x^{\infty}$", which is uniformly zero in $(-1,1)$.
 A: The counterexample given here was suggested by @Nate Eldredge in the comments; I'm just elaborating on its properties :) 

The function $f: \Bbb{R} \to \Bbb{R}$ defined by
\begin{align}
f(x) = 
\begin{cases}
e^{-1/x^2} \sin\left( \frac{1}{x}\right) & \text{if $x \neq 0$} \\
0 & \text{if $x=0$}
\end{cases}
\end{align}
is easily seen to be $C^{\infty}$ away from the origin, and at the origin, one can show that all the derivatives vanish. The most straight-forward proof I know is by direct verification (a really strict proof follows by induction on the form of the derivative).
The rapid oscillatory behaviour of $f$ near the origin shows that there is no $\varepsilon > 0$ for which those conditions you stated hold. (I suggest you use wolfram alpha to plot this function to see just how quickly things approach $0$ at the origin, and how fast the function is oscillating).

Here's a rough idea of the proof of $C^{\infty}$ at the origin. Let's first show that $f'(0)$ exists and equals $0$. For $x\neq 0$, we have that
\begin{align}
 \left | \dfrac{f(0 + x) - f(0)}{x} \right| &= \left| \dfrac{e^{-1/x^2} \sin (1/x)}{x} \right| \\
& \leq \left| \dfrac{e^{-1/x^2}}{x} \right| \cdot 1
\end{align}
And, you should know from somewhere that "exponentials dominate polynomials", in the sense that the numerator goes to $0$ much faster than the denominator goes to $\pm\infty$, so as $x \to 0$, the RHS tends to $0$ as well.
In general, you can show that for $x \neq 0$, the $k^{th}$ derivative looks like an exponential term multiplied by trigonometric term multiplied by a polynomial in $\dfrac{1}{x}$. I.e There exist polynomials $P,Q$ such that
\begin{align}
f^{(k)}(x) = e^{-1/x^2} \left( P\left(\dfrac{1}{x} \right) \cdot \sin\left(\dfrac{1}{x} \right) + Q\left(\dfrac{1}{x} \right) \cdot \cos \left(\dfrac{1}{x} \right)\right)
\end{align}
And as $x \to 0$, the limit will be $0$, because "exponentials dominate polynomials" (the trigonometric terms are bounded by $1$; so they don't really matter). 
