# Does this prove the Squeeze Theorem?

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

• You are using the squeeze theorem in your proof of the squeeze theorem. You should use the definition of limit, instead. – Giuseppe Negro Oct 9 '18 at 8:00
• assuming the validity of something you try to prove is never fruitful – Alvin Lepik Oct 9 '18 at 8:06
• 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. – 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?

• 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. – Martin Rosenau Oct 9 '18 at 13:53
• ... sorry, I wanted to write $y_n=sin(\frac{n\pi}{2})$; $y_n=sin(n\pi)$ is constantly 0. – 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

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

Can you proceed ?

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