When can $f_1(x)\cos(\phi(x))+f_2(x)\sin(\phi(x))$ have a limit when $\phi(x)$ tends to infinity? Let $f_1,f_2:[0,\infty) \to \mathbb R$ be smooth functions, and let $\phi:(0,\infty) \to \mathbb R$ be a smooth function satisfying $\lim_{x \to 0^+}\phi(x)=\infty$ or $\lim_{x \to 0^+}\phi(x)=-\infty$. Note that $\phi$ is not defined at $x=0$.

Suppose that $ \lim_{x \to 0^+} f_1(x)\cos(\phi(x))+f_2(x)\sin(\phi(x)) $
  exists. Is it true that $ f_1(0)=f_2(0)=0$? 

The assumptions imply that both limits $a_i:=\lim_{x \to 0^+} f_i(x)=f_i(0)$ exist. The question is then wether we must have $a_1=a_2=0$.
If one of the $f_i$ tends to zero when $x \to 0$, then so must the other one- since  $\sin(\phi(x))$ and $\cos(\phi(x))$ by themselves don't converge to anything. 
We somehow need to exclude the possibility that both $f_i$ have non-zero limits, and the two summands somehow "cancel" each other nicely. 
 A: Let us first consider the easy case when $f_1,f_2$ are locally constant around $0$. The limit $$\lim\limits_{x \to 0^+} f_1(x)\cos(\phi(x))+f_2(x)\sin(\phi(x))$$ reduces to $$\lim\limits_{x \to 0^+} a\cos(\phi(x))+b\sin(\phi(x))$$ where $a=f_1(0),b=f_2(0)$.
As we move $x$ closer to $0$, sometimes we get $\cos(\phi(x))=0$ and $\sin(\phi(x))=1$, while sometimes we get $\cos(\phi(x))=1$ and $\sin(\phi(x))=0$. The limit is actually oscillating between $a$ and $b$ (and many other numbers but those are not important). But we are saying the limit exists, so it must be that $a=b$.
The limit is now equal to $$\lim\limits_{x \to 0^+} a(\cos(\phi(x))+\sin(\phi(x))).$$ This limit is still oscillating and does not exist, unless $a=0$. So we have $f_1(0)=f_2(0)=0$, if they are locally constant.
What about the general case when they are not locally constant? Easy: consider $f_1(x)=a+\varepsilon_1(x)$, $f_2(x)=b+\varepsilon_2(x)$, where $a=f_1(0),b=f_2(0)$. The two functions $\varepsilon_i$ tend to $0$ around $0$. Using squeeze theorem, the limit $$\lim\limits_{x \to 0^+} \varepsilon_1(x)\cos(\phi(x))+\varepsilon_2(x)\sin(\phi(x))$$ exists and is zero. So the limit $$\lim\limits_{x \to 0^+} a\cos(\phi(x))+b\sin(\phi(x))$$ also exists and is equal to $$\lim\limits_{x \to 0^+} f_1(x)\cos(\phi(x))+f_2(x)\sin(\phi(x))-\lim\limits_{y \to 0^+} \varepsilon_1(y)\cos(\phi(y))+\varepsilon_2(y)\sin(\phi(y)).$$ From previous arguments,we must have $a=b=0$.
A: After some more thought, I think that the answer is that we must have $f_i(0)=0$. Is there a simpler solution?
Indeed, since $\lim_{x \to 0^+}\phi(x)=\pm \infty$ there exist sequences $x_n,y_n \to 0$ such that
$$ \cos(\phi(x_n))=1, \sin(\phi(x_n))=0, $$
and
$$ \cos(\phi(y_n))=0, \sin(\phi(y_n))=1. $$
The assumption that $L:=\lim_{x \to 0^+} f_1(x)\cos(\phi(x))+f_2(x)\sin(\phi(x))$ exists implies that
$f_1(0)=\lim_{n \to \infty} f_1(x_n)=L=\lim_{n \to \infty} f_2(y_n)=f_2(0)$, so $$f_1(0)=f_2(0)=L.$$
Now, let $\theta \in [0,2\pi]$ be an arbitrary angle. There exist  a sequence $z_n \to 0$ such that $\phi(z_n)=\theta \mod 2\pi\mathbb Z$, so
$$ \cos(\phi(z_n))=\cos \theta, \sin(\phi(z_n))= \sin \theta. $$ 
This implies that 
$$L=\lim_{n \to \infty} f_1(z_n)\cos(\phi(z_n))+f_2(z_n)\sin(\phi(z_n))=\\ f_1(0) \cos \theta+f_2(0) \cos \theta=L(\cos \theta+\sin \theta).$$
This forces $L=0$; if $L \neq 0$, then we could conclude that $\cos \theta+\sin \theta=1$ for any angle $\theta \in [0,2\pi]$, which is a contradiction.
