Squeeze theorem

Suppose $b_n\le a_n\le c_n$ for all sufficiently large $n$, and $\lim b_n=\lim c_n=L$.

Fix $\varepsilon>0$. Since $\lim b_n=L$, there is $N_1$ with $L-\varepsilon<b_n<L+\varepsilon$ for all $n>N_1$. Since $\lim c_n=L$, there is $N_2$ with $L-\varepsilon<c_n<L+\varepsilon$ for all $n>N_2$.

Let $N=\max(N_1,N_2)$. For every $n>N$, combining $b_n\le a_n\le c_n$ with both inequalities above gives $L-\varepsilon<b_n\le a_n\le c_n<L+\varepsilon$, hence $L-\varepsilon<a_n<L+\varepsilon$, i.e. $|a_n-L|<\varepsilon$.

Since $\varepsilon>0$ was arbitrary, $\lim_{n\to\infty} a_n=L$ by the ε–N definition. $\blacksquare$