= Solution
{c}
If $L<1$, set $\varepsilon=(1-L)/2>0$. The definition of <limit of a sequence> gives $a_n<L+\varepsilon=(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\geq1$. Hence $\boxed{L\geq1}$.
Factor the difference of cubes. Since both <cube roots> are at least one,
$$
\left|a_n^{1/3}-L^{1/3}\right|
=\frac{|a_n-L|}{a_n^{2/3}+a_n^{1/3}L^{1/3}+L^{2/3}}
\leq\frac{|a_n-L|}{3}.
$$
Given $\varepsilon>0$, the definition of <limit of a sequence> supplies $N$ with $|a_n-L|<3\varepsilon$ for $n\geq N$. The displayed <inequality> then gives $|a_n^{1/3}-L^{1/3}|<\varepsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.
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