TITLE: How to prove the Squeeze Theorem for sequences
QUESTION [18 upvotes]: The formulation I'm looking at goes: If $\lbrace x_n\rbrace$, $\lbrace y_n\rbrace$ and $\lbrace z_n \rbrace$ are sequences such that $x_n \le y_n \le z_n$ for all $n \in \mathbb N$, and $x_n \to l$ and $z_n \to l$ for some $l \in \mathbb R$, then $y_n \to l$ also.
So we have to use the definition of convergence to a limit for a sequence: 
$$\forall \varepsilon > 0, \space \exists N_\varepsilon \in \mathbb N, \space \forall n \ge N_\varepsilon, \space |a_n - l| < \varepsilon$$
I've been trying to say something like: $|y_n - l| < |x_n - l| + |z_n - l| \le \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon$ for every $\varepsilon > 0$, but I'm not sure how to get there or if there may be a better way to prove the theorem. Any help would be greatly appreciated.

REPLY [3 votes]: An alternative proof is the following:  
Proof: Since $x_n \leq y_n \leq z_n$ then $0\leq y_n-x_n\leq z_n-x_n$, thus $|y_n-x_n|\leq z_n-x_n$. 
Combining the above with the fact that $\lim(z_n-x_n)=\lim z_n-\lim x_n=l-l=0 \ $,  we get: $$\lim(y_n-x_n)=0$$ 
Now we can write the terms of $(y_n)$ as the sum of the terms of two converging sequences: $y_n=(y_n-x_n)+x_n$, so we have:
$$
\lim y_n=\lim\big((y_n-x_n)+x_n \big)=\lim(y_n-x_n)+\lim x_n=0+l=l
$$