I'm aware that there are plenty of proofs for the squeeze theorem but I wanted to verify if I'm on the right track for this approach.

Problem: Show that if $x_n \leq y_n \leq z_n \hspace{1mm} \forall n \in \mathbf{N}$ and if lim$(x_n)$ = lim$(z_n) = \ell$, then lim$(y_n) = \ell$ as well.

Solution: Given that $x_n \leq y_n \leq z_n \hspace{1mm} \forall n \in \mathbf{N}$ then $y_n$ converges otherwise $x_n$ or $z_n$ diverge. Let lim$(y_n) = y$. Then,

\begin{align} &x_n \leq y_n \leq z_n \\ &0 \leq y_n-x_n \leq z_n-x_n \\ \implies &\text{lim}(y_n-x_n) = y-\ell \leq \text{lim}(z_n-x_n) = \ell- \ell = 0 \\ \implies &y \leq \ell \end{align}

Similarly, \begin{align} &x_n-z_n \leq y_n-z_n\leq 0 \\ \implies&\text{lim}(x_n-z_n) = \ell-\ell = 0 \leq \text{lim}(y_n-z_n)=y-\ell \\ \implies&\ell \leq y \end{align}

Hence, $y=\ell \hspace{1cm}\square$

Okay so I used @Fred's hint and changed the proof.

$\epsilon < x_n-\ell \leq y_n-\ell \leq z_n -\ell < \epsilon$

since lim$(x_n)$ = lim$(z_n)= \ell$

Thus $-\epsilon < y_n - \ell < \epsilon \implies |y_n-\ell|<\epsilon$

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    $\begingroup$ You are using the squeeze theorem in your proof of the squeeze theorem. You should use the definition of limit, instead. $\endgroup$ – Giuseppe Negro Oct 9 '18 at 8:00
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    $\begingroup$ assuming the validity of something you try to prove is never fruitful $\endgroup$ – Alvin Lepik Oct 9 '18 at 8:06
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    $\begingroup$ I just want to spell something out which is implicit in the comments and answers. The conclusion of the squeeze theorem is really two parts. First, it is that $\lim y_n$ exists. Second, its is that $\lim y_n = \ell$. Your proof assumes the first conclusion and proves the second. $\endgroup$ – Jason DeVito Oct 9 '18 at 19:43

Given that $x_n \leq y_n \leq z_n \hspace{1mm} \forall n \in \mathbf{N}$ then $y_n$ converges otherwise $x_n$ or $z_n$ diverge.

How do you know this?

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    $\begingroup$ There is even a counterexample: $x_n=-2$, $y_n=sin(n\pi)$, $z_n=2$. $y_n$ does not converge although $x_n$ and $z_n$ do. $\endgroup$ – Martin Rosenau Oct 9 '18 at 13:53
  • $\begingroup$ ... sorry, I wanted to write $y_n=sin(\frac{n\pi}{2})$; $y_n=sin(n\pi)$ is constantly 0. $\endgroup$ – Martin Rosenau Oct 9 '18 at 20:22

It is wrong from the start. You are supposed to prove that, since $(x_n)_{n\in\mathbb N}$ and $(z_n)_{n\in\mathbb N}$, then $(y_n)_{n\in\mathbb N}$ converges. And the first thing that you do is to assert that $(y_n)_{n\in\mathbb N}$ converges.


If $\epsilon >0$, then there is $N \in \mathbb N$ such that

$\ell- \epsilon <x_n$ and $z_n < \ell + \epsilon$ for all $n>N$.

Can you proceed ?

  • $\begingroup$ Ah that makes sense because then -epsilon < yn - \ell < epsilon so |yn-\ell| < epsilon $\endgroup$ – ABC Oct 9 '18 at 8:07
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    $\begingroup$ OP is aware that there are alternative proof.... $\endgroup$ – user202729 Oct 9 '18 at 11:35

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