Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 2 / a / i / Solution

If $L<1$, set $\epsilon =(1-L)/2>0$. The definition of limit of a sequence gives $a_n<L+\epsilon =(1+L)/2<1$ for all sufficiently large $n$, contradicting $a_n\ge 1$. Hence $\boxed{L\ge 1}$.

Factor the difference of cubes. Since both cube roots are at least one,Given $\epsilon >0$, the definition of limit of a sequence supplies $N$ with $|a_n-L|<3\epsilon$ for $n\ge N$. The displayed inequality then gives $|a_n^{1/3}-L^{1/3}|<\epsilon$, proving directly that $\boxed{a_n^{1/3}\to L^{1/3}}$.