= Solution
Set $L=\inf_{k\ge1}x_k/k$. It may be $-\infty$, but cannot be $+\infty$ because $x_1$ is finite. Fix $k$ and write $n=qk+r$, $0\le r<k$. Repeated use of the <subadditive sequence> inequality gives
$$
x_n\le qx_k+C_k,\qquad C_k=\max(0,x_1,\ldots,x_{k-1}),
$$
where $C_1=0$. Dividing by $n$ and letting $n\to\infty$ gives $\limsup x_n/n\le x_k/k$. Since this holds for every $k$, the upper <limit of a sequence> is at most $L$, while every ratio is at least $L$. If $L$ is finite these bounds agree. If $L=-\infty$, choose $k$ with $x_k/k$ below any prescribed real number; the same upper-limit bound proves convergence to $-\infty$. This establishes the <extended-real Fekete lemma>:
$$
\boxed{\lim_{n\to\infty}\frac{x_n}{n}=\inf_{k\ge1}\frac{x_k}{k}\in[-\infty,\infty).}
$$
A finite real <limit of a sequence> requires a lower linear bound. For example $x_n=-n^2$ satisfies the premise but has normalized <limit of a sequence> $-\infty$, so that qualification cannot be dropped.
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