How to prove the Squeeze Theorem for sequences

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.

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Write

$$|y_n-l|\le |y_n-x_n|+|x_n-l|\le( z_n-x_n)+|x_n-l|$$
Can you take it from here?

Let $\varepsilon > 0$. Since $x_n \to l$, there exists $N_1 = N_1(\varepsilon)$ such that $|x_n – l| < \varepsilon$ for all $n \ge N_1$. Since $z_n \to l$, there exists $N_2 = N_2(\varepsilon)$ such that $|z_n – l| < \varepsilon$ for all $n \ge N_2$. Set $N = \max\{N_1,N_2\}$. If $n \ge N$, then $$y_n – l \le z_n – l < \varepsilon$$ and $$y_n – l \ge x_n – l > -\varepsilon$$ Hence $|y_n – l| < \varepsilon$ for all $n \ge N$. Since $\varepsilon$ was arbitrary, $y_n \to l$.

Ok, so this is what I have now:

Let $x_n \le y_n \le z_n \space\space \forall n \in \mathbb N$ and $\lim_{n \to \infty} x_n = \lim_{n \to \infty} z_n = l$ for some $l \in \mathbb R$.

Then we have:
$$\begin{align}
&\forall \varepsilon_1 > 0,\, \exists N_{\varepsilon_1} \in \mathbb N,\, \forall n \ge N_{\varepsilon_1}\space |x_n – l| < \varepsilon_1 \\
&\forall \varepsilon_2 > 0,\, \exists N_{\varepsilon_2} \in \mathbb N,\, \forall n \ge N_{\varepsilon_2}\space |z_n -l| < \varepsilon_2
\end{align}$$
Let $N = \max\lbrace N_{\varepsilon_1},N_{\varepsilon_2} \rbrace$ and $\varepsilon = \min\lbrace \varepsilon_1, \varepsilon_2 \rbrace$, so that we have
$$\forall \varepsilon > 0,\, \exists N \in \mathbb N,\, \forall n \ge N\space |x_n – l| < \varepsilon \space \text{and} \space |z_n -l| < \varepsilon$$

Let $\frac\varepsilon3 > 0$. Then $\exists N \in \mathbb N,\, \forall n \ge N\space |x_n -l| < \frac\varepsilon3 \text{ and } |z_n -l| < \frac\varepsilon3$ so that
$$|z_n – x_n| = |z_n – l + l – x_n| \le |z_n – l| + |x_n – l| < \frac\varepsilon3 + \frac\varepsilon3 = \frac{2\varepsilon}3$$
By assumption,
$$\begin{align}
x_n \le&\, y_n \le z_n \\
0 \le&\, y_n – x_n \le z_n – x_n \\
&|y_n – x_n| \le |z_n – x_n| \\
|y_n – l| = |y_n – x_n + x_n – l| \le& |y_n – x_n| + |x_n – l| \le |z_n – x_n| + |x_n – l| < \frac{2\varepsilon}3 + \frac\varepsilon3 = \varepsilon
\end{align}
$$
And so $\lbrace y_n \rbrace$ also converges to $l$, thus, Q.E.D.

I think that that’s satisfactory. If there are any ambiguities or mistakes, please let me know. Thanks to everybody for your advice and contributions.

How about
$$
|y_n-l|\leq\max\{|x_n-l|,|z_n-l|\}<\varepsilon
$$
where last inequality of the above holds for $n\geq N$ if $N$ is big enough?

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:
$$
\lim y_n=\lim\big((y_n-x_n)+x_n \big)=\lim(y_n-x_n)+\lim x_n=0+l=l
$$

As $n$ grows, you can get the distance from $x_n$ to $l$ to be less than any $\epsilon > 0$, and the distance from $z_n$ to $l$ to be less than $\epsilon$ as well. Just take the max $N$ of the two indices for $x_n$ and $z_n$ that guarantee this, and you will get that both $x_n$ and $z_n$ are within $\epsilon$ of $l$ for $n \geq N$. What does that say about the distance between $y_n$ and $l$, for $n \geq N$?