# Proof of the Squeeze theorem for Sequences

Show that if $$x_n \le y_n \le z_n$$ for all $$n \in \mathbb N$$, and if $$lim(x_n) = lim (z_n) = l$$, then $$lim (y_n)= l$$ as well.

Proof. Since $$lim(x_n) = lim (z_n) = l$$. Then follows that $$|x_n -l| < \frac{\epsilon}{2}$$ whenever $$n \ge N_1$$ and similarly $$|l-z_n | < \frac{\epsilon}{2}$$ whenever $$n \ge N_2$$. Choose $$max \{ N_1,N_2\}$$ and by the triangle inequality we get $$|x_n - z_n| \le|x_n -l|+ |l-z_n| \lt \epsilon .$$

since $$|x_n - z_n| \lt \epsilon$$ it follows that $$x_n=z_n$$ for sufficiently large $$n \ge \max\{N_1, N_2\}$$. Hence it follows that $$x_n=y_n= z_n$$ (from the fact that $$x_n=z_n$$ and $$x_n \le y_n \le z_n$$). And therefore $$lim(x_n) = lim (y_n) = l$$.

Question 1: Is my attempt at the proof correct? If not why not? Also if not can you give a correct version or atleast a few hints?

Question 2: I use the property that $$|a-b|< \epsilon$$ is equivalent to $$a=b$$ but it doesn't sit well with me below that $$x_n=z_n$$ since $$(x_n), (z_n)$$ could have different elements and still converge at the same value.) Why am I getting contradictory results here?

I really appreciate this site helping me learn how to write proofs in real analysis :)

Firstly, $z_n$ does not equal $x_n$ for sufficiently large $n$. For an answer to your second question $\frac{1}{n}$ and $\frac{1}{2n}$ both converge to zero, but they are never equal for a given $n$. Use the definition of convergence instead. What do you need to show that $\lim_{n\to \infty} y_n=l$?

• Yes that is correct – Red Jul 15 '18 at 2:55
• I completely agree with you that $z_n$ does not necessarily equal to $x_n$ but as I have shown the inequality $|x_n - z_n| \lt \epsilon$ would imply that $x_n = z_n$ doesn't it?? – Red Jul 15 '18 at 2:59
• @Red: why would the inequality imply $z_n=x_n$? – Paramanand Singh Jul 15 '18 at 6:06

Most elementary results of convergence for sequences either depend on the triangle inequality (via an $\epsilon/2$ trick) or the inequality $0\leq\frac{|x|}{1+|x|}\leq 1$ for all $x$ (to avoid cases where division by $0$ may occur).

The Squeeze Theorem however is probably best proved by a different approach.

Try to show that if $x\leq y\leq z$ then $|y-l|\leq \max\left\{|x-l|, |z-l|\right\}$ no matter what $l$ may be. This is probably best proved with a picture.

• Is the aproach I have taken completely wrong or can it be tweaked for a correct proof? – Red Jul 15 '18 at 3:11
• @Red you have claimed that the sequences $\{x_n\}$ and $\{z_n\}$ eventually agree. But an easy counter example is $x_n=-1/n$, $z_n=1/n$ and $y_n=0$ for all $n$. – Robert Wolfe Jul 15 '18 at 3:12
• I completely agree with this, the sequences need not agree eventually.But why is the triangle inequality gving me that result anyway? I feel like I have used it wrong but I cannot figure out where my mistake llies? as I said in th second part of my question – Red Jul 15 '18 at 3:16
• @Red if $a$ and $b$ satisfy $|a-b|<\epsilon$ for every $\epsilon>0$ then $a=b$. However you have not proved that $|x_n-z_n|<\epsilon$ for every $\epsilon>0$. You have proved that for every $\epsilon>0$ there is an $n$ such that $|x_n-z_n|<\epsilon$. – Robert Wolfe Jul 15 '18 at 3:27
• I see my mistake. Thanks! – Red Jul 15 '18 at 3:28