= Solution
For fixed $n$, define the remainders
$$
r_n=f-\sum_{j=0}^na_j\phi_j,
\qquad
r_{n+1}=f-\sum_{j=0}^{n+1}a_j\phi_j.
$$
The hypotheses say $r_n=o(\phi_n)$ and $r_{n+1}=o(\phi_{n+1})$, while subtraction gives
$$
r_n=a_{n+1}\phi_{n+1}+r_{n+1}.
$$
Because $a_{n+1}\ne0$,
$$
\frac{r_n}{\phi_{n+1}}\longrightarrow a_{n+1}.
$$
For $x$ sufficiently close to $x_0$ this denominator is therefore nonzero, and
$$
\frac{\phi_{n+1}}{\phi_n}
=\frac{r_n/\phi_n}{r_n/\phi_{n+1}}
\longrightarrow\frac0{a_{n+1}}=0.
$$
\b[Thus $\phi_{n+1}=o(\phi_n)$ for every $n$, so $(\phi_n)$ is an <asymptotic sequence>.]
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