of [[Convergence]]. **Lemma** Suppose there exists $N \in \mathbb{N}$ such that for all $n \geq N$ $a_{n} \leq b_{n} \leq c_{n} $and $a_{n} \to l$ and $c_{n} \to l$. Then $b_{n} \to l$. **Proof** Since $a_{n} \to l$ there exists $N_{1}$ such that for all $n \geq N_{1}$, $l-\epsilon<a_{n}<l+\epsilon$since $c_{n} \to l$ there exists $N_{2}$ such that for all $n \geq N_{2}$. $l-\epsilon<c_{n}<l+\epsilon$If we take take $n \geq N:= \max(N_{1},N_{2})$ then $l-\epsilon< a_{n} \leq b_{n} \leq c_{n} < l+ \epsilon$i.e. $|b_{n}-l|<\epsilon$ which shows that $b_{n} \to l$. ### Examples - [[x to the power of the reciprocal of n converges to 1 for any positive x]]. - [[n to the power of the reciprocal of n converges to 1]].