Sandwich theorem

Definition: Sandwich theorem

The sandwich theorem (also called squeeze theorem) says that if a sequence is trapped between two other sequences with the same limit, then it has that limit as well.

Theorem

Let $(a_n),(b_n),(c_n)$ be real sequences and assume $a_n\le b_n\le c_n$ for all $n\ge N_0$ for some $N_0\in \mathbb{N}$. If ${\lim }_{n\to \infty }{a}_n=L\text{and}{\lim }_{n\to \infty }{c}_n=L,$ then ${\lim }_{n\to \infty }{b}_n=L.$

Proof idea

Take any $\epsilon >0$. Since $a_n\to L$, there is $N_1$ such that $|a_n-L|<\epsilon$ for all $n>N_1$, so $L-\epsilon <a_n$. Since $c_n\to L$, there is $N_2$ such that $|c_n-L|<\epsilon$ for all $n>N_2$, so $c_n<L+\epsilon$. For $n>\max \{N_0,N_1,N_2\}$: $L-\epsilon <a_n\le b_n\le c_n<L+\epsilon ,$ therefore $|b_n-L|<\epsilon$, which is exactly convergence.

Example

Show that $b_n=\frac{\sin n}{n}$ converges to $0$. Since $-1\le \sin n\le 1$, dividing by $n>0$ gives $-\frac{1}{n}\le \frac{\sin n}{n}\le \frac{1}{n}.$ Both bounds converge to $0$, so by the sandwich theorem, ${\lim }_{n\to \infty }\frac{\sin n}{n}=0.$