Implications of differentiability and Taylor expansion Consider a function $\phi: \Theta \subseteq \mathbb{R}^l \rightarrow \mathbb{R}$. Fix $\theta_0 \in \Theta$.
Assume:
(1) $\phi(\cdot)$ differentiable at $\theta_0$
(2)  The gradient at $\theta_0$, $\dot{\phi}_{\theta_0}$, is different from zero
(3) $\phi(\theta_0)=0$
Let 
(i) $\{r_n\}_n$ be a sequence of real numbers such that $\lim_{n\rightarrow \infty}r_n=\infty$
(ii) $(\theta_0+\frac{h}{r_n}) \in \Theta$, with $h \in \mathbb{R}^l$ such that $\dot{\phi}_{\theta_0}h<0$
If I take the first order Taylor Expansion of $\phi(\theta)$ around $\theta_0$ and I evaluate it at $\theta_o+\frac{h}{r_n}$ I get
$$
\phi(\theta_0+\frac{h}{r_n})=\frac{\dot{\phi}_{\theta_0}h}{r_n}+o(1)
$$
as $n\rightarrow \infty$.
Can I conclude $\phi(\theta_0+\frac{h}{r_n})<0$ for all $n$ or for $n$ large only?
(from van der Vaart "Asymptotic Statistics" proof Theorem 15.4)
 A: Yes you can conclude that $\phi(\theta_0+\frac{h}{r_n})<0$ for $n \to +\infty$.
By using your equation with a $\frac{o(1)}{r_n}$ instead of the $o(1$) (and what @zhw,@STF said), you can prove it quickly:
Assume $\phi(\theta_0+\frac{h}{r_n}) \geq 0$ for all $n$, then the limit gives $\frac{\dot{\phi}_{\theta_0}h}{r_n} \to 0^{+}$, absurd because they are all $< 0$.
A: You are using differentiability in the following form: If $\phi:\>{\mathbb R}^l\to{\mathbb R}$ is diferentiable at $0$, and $\phi(0)=0$, then
$$\phi(x)=\nabla \phi(0)\cdot x+o\bigl(|x|\bigr)\qquad(x\to0)\ ,$$ which is a condensed way of saying that the truncation error
$$\rho(x):=\phi(x)-\nabla \phi(0)\cdot x$$
satisfies
$$\lim_{x\to0}{\rho(x)\over|x|}=0\ .\tag{1}$$
You now choose a unit vector $u\in{\mathbb R}^l$ and consider the vectors $x_n:=\epsilon_n\,u$ for some sequence $(\epsilon_n)_{n\geq1}$ of positive numbers converging to $0$. It follows that $|x_n|=\epsilon_n$. We therefore can write
$$\phi(x_n)=\nabla \phi(0)\cdot (\epsilon_n u) +\rho(\epsilon_n u)=\epsilon_n\left(\nabla\phi(0)\cdot u+{\rho(x_n )\over|x_n|}\right)\ .$$
Suppose now that $\nabla\phi(0)\cdot u\ne0$. From $(1)$ we then can conclude that there is an $n_0$ such that 
$${\bigl|\rho(x_n )\bigr|\over|x_n|}<\bigl|\nabla\phi(0)\cdot u\bigr|\qquad(n>n_0)\ .$$
Only for these "large" $n$ we can guarantee that
$${\rm sgn}\bigl(\phi(x_n)\bigr)={\rm sgn}\bigl(\nabla\phi(0)\cdot u\bigr)\ .$$
Consider the following example with $l=1$:
$$\phi(x):=x-4|x|^{3/2};\qquad u:=1, \quad \epsilon_n:={1\over n}\ (n\geq1)\ .$$
Then $\phi$ is differentiable at $0$, $\ \phi'(0)\cdot u=1>0$, but
$$\phi(x_n)={1\over n}\left(1-4\sqrt{{1\over n}}\right)$$
is negative for all $n<16$.
