= Solution
First suppose $x_n\to L>0$. Continuity of the <natural logarithm> gives $\log x_n\to\log L$. Part (a), applied to this sequence, yields
$$
\frac1n\sum_{k=1}^n\log x_k\longrightarrow\log L.
$$
Applying the continuous <exponential function>,
$$
\sqrt[n]{x_1x_2\cdots x_n}
=\exp\!\left(\frac1n\sum_{k=1}^n\log x_k\right)
\longrightarrow L.
$$
If $L=0$, then for every $\varepsilon>0$ all sufficiently late $x_k$ are below $\varepsilon$. Splitting off the fixed initial product shows that the limsup of the geometric means is at most $\varepsilon$; hence it is zero. This proves the <geometric mean of a sequence> result in all cases.
Now suppose
$$
r_n=\frac{x_n}{x_{n-1}}\longrightarrow r.
$$
The telescoping product is
$$
x_n=x_0\prod_{k=1}^nr_k.
$$
Therefore
$$
\sqrt[n]{x_n}
=x_0^{1/n}\sqrt[n]{r_1r_2\cdots r_n}
\longrightarrow 1\cdot r,
$$
and so
$$
\boxed{\lim_{n\to\infty}\sqrt[n]{x_n}
=\lim_{n\to\infty}\frac{x_n}{x_{n-1}}}.
$$
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