**Proof Plan.** The hypothesis $a_n \le b_n \le c_n$ combined with $a_n \to L$ and $c_n \to L$ traps $b_n$ in an interval of width $|c_n - a_n| \to 0$ around $L$. **Step 1 (Set up the $\varepsilon$-bound).** Let $\varepsilon > 0$. Since $a_n \to L$, there exists $N_1$ with $|a_n - L| < \varepsilon$ for all $n \ge N_1$, i.e. $L - \varepsilon < a_n$. Since $c_n \to L$, there exists $N_2$ with $|c_n - L| < \varepsilon$ for all $n \ge N_2$, i.e. $c_n < L + \varepsilon$. **Step 2 (Conclude).** Set $N = \max(N_1, N_2)$. For $n \ge N$: \begin{align*} L - \varepsilon < a_n \le b_n \le c_n < L + \varepsilon, \end{align*} so $|b_n - L| < \varepsilon$. Since $\varepsilon > 0$ was arbitrary, $b_n \to L$. $\blacksquare$