# Can I prove squeeze theorem this way?

The question is: Show that if $$x_n \leq y_n \leq z_n$$ for all $$n \in \mathbb N$$, and if $$lim x_n = limz_n = l$$, then $$limy_n =l$$ as well.

So far for my solution I have: Using the definition of convergence, we know $$|x_n - l| < \epsilon$$ for some $$\epsilon > 0$$, and $$|z_n - l| < \epsilon$$ for some $$\epsilon > 0$$.

Using this, we have $$|x_n - l| - |z_n - l| < \epsilon - \epsilon$$ which goes to $$|x_n - l - z_n + l| < 0$$, thus $$|x_n - z_n| < 0$$.

We know absolute value cant be less than 0, but I'm not sure where I could incorporate a $$\leq$$ sign so that the final inequality becomes $$|x_n - z_n| \leq 0$$, proving that $$x_n = z_n$$, therefor $$x_n = y_n = z_n$$, so $$limy_n = l$$ as well. Can I do this, or am I going in the wrong direction completely? Thanks!

• u shouldn't be able to prove $x_n = z_n$ because it's not always true. There's a mistake here: $|x_n - l | < \epsilon$, $|z_n - l| < \epsilon$ doesn't mean $|x_n - l | - |z_n - l| < 0$, you can't "substract" same-sign inequalities, only add them. – Dominik Kutek Oct 26 '19 at 18:54
• If $a < \epsilon$ and $b < \epsilon$ does not mean that $a -b < \epsilon - \epsilon$. – copper.hat Oct 26 '19 at 18:55
• Furthermore the step where you merge the two absolute values into one is also incorrect; this isn't always true. Example: $|(-3)-0| - |6-0| = -3 \neq |(-3)-0-6+0|$ – WaveX Oct 26 '19 at 19:05

No that's not a proper way, indeed for example

$$(3<4) \quad \land \quad (2<4) \quad \not\Rightarrow \quad (3-2)<0$$

That is wrong, for several reasons:

• Asserting that $$\lim_{x\to\infty}x_n=\lim_{n\to\infty}z_n=l$$ does not meant that $$\lvert x_n-l\rvert,\lvert z_n-l\rvert<\varepsilon$$ for some $$\varepsilon>0$$; it means that you have that inequality for every $$\varepsilon>0$$, if $$n$$ is large enough.
• There is no way you can prove that $$\lvert x_n-l-z_n+l\rvert<0$$; an absolute value is always greater than or equal to $$0$$.

You can prove it as follows. Take $$\varepsilon>0$$. If $$n$$ is large enough, then $$\lvert x_n-l\rvert,\lvert z_n-l\rvert<\varepsilon$$, which means that $$x_n,z_n\in(l-\varepsilon,l+\varepsilon)$$. But then, since $$x_n\leqslant y_n\leqslant z_n$$, $$y_n\in(l-\varepsilon,l+\varepsilon)$$, which means that $$\lvert y_n-l\rvert<\varepsilon$$ (if $$n$$ is large enough).

Jose has explained the two errors in your attempt. Here is a hint to get you on the right track.

Fix $$\epsilon > 0$$. Your goal is to show that for all large $$n$$ you have $$|y_n - l| < \epsilon$$.

The key observation is $$|y_n - l| \le \max\{|x_n - l|, |z_n - l|\}$$. (For instance, if $$y_n \le l$$ then $$|y_n - l| \le |x_n - l|$$.) Now try to show that the right-hand side is smaller than $$\epsilon$$ for all large $$n$$, using the fact that $$\lim_{n \to \infty} x_n = \lim_{n \to \infty}z_n = l$$.

Using the definition of convergence, we know $$|x_n−l|<ϵ$$ for some $$ϵ>0$$.

That is, sadly, not the definition for convergence and is actually a pretty serious botch.

If $$x_n= \frac 1{2^n}$$ then what is $$\lim x_n$$. Well by your definition, $$0 < x_n \le 1$$ so $$-39 < x_n - 39 \le -38$$ so $$|x_n-39| < 39$$. So for $$\epsilon = 39$$ we have $$|x_n -39| < \epsilon$$ so $$\lim x_n = 39$$.

The actual definition is that for any $$\epsilon > 0$$ we can find some $$N$$ so that whenever $$n > N$$ we will have $$|x_n -l| < \epsilon$$.

"Using the definition of convergence, we know |xn−l|<ϵ for some ϵ>0, and |zn−l|<ϵ for some ϵ>0"

we should say:

Using the definition of convergence, we know that for any $$\epsilon > 0$$ there are $$N_1$$ and $$N_2$$ so that for all $$n > N_1$$ we'd have $$|x_n−l|<ϵ$$, and for all $$n > N_2$$ we'd have $$|z_n−l|<ϵ$$. ANd for all $$n > \max(N_1, N_2)$$ we'd have both $$|x_n - l|< \epsilon$$ and $$|z_n -l|< \epsilon$$.

Now you claim that $$|x_n-l|< e$$ and $$|z_n-l| < e$$ implies $$|x_n-1|-|z_n-l| < e-e$$. This is utterly wrong. Negatives flip the inequalities so $$-|z_n - l| > -\epsilon$$. And if you have $$|x_n-1|$$ is less then $$\epsilon$$. And $$-|z_n-l| > -\epsilon$$ you can't say any thing about how they combine.

For example $$4 < 5$$ and $$1 < 4$$ so $$4- 1 < 5-4$$. Really?

But what you can say is $$|x_n -l| + |z_n-l| < \epsilon + \epsilon$$.

The you claim that $$|x_n-l| - |z_n -l| = |(x_n -l)-(z_n-1)|$$. You can not combine absolute values that way. Consider $$|5|-|-3| = 5 - 3 =2$$ and $$|(5)-(-3)| = |5+3| = 8$$.

What you need to do is use addition and the triangle inequality: $$|a-b| + |b-c| \ge |a-c|$$.

so $$|(x_n-l)+ (l- z_n)| \le |x_n-l|+|z_n - 1| < \epsilon + \epsilon = 2\epsilon$$.

How do we put this all together?

........

We want to find an $$\mathscr N$$ so that for all $$n >\mathscr N$$ we have $$|y_n - l|\epsilon$$.

And we know we can talk about $$|z_n - l|$$ and $$|y_n - l|$$.

So $$|y_n - l|= |(y_n - x_n) +(x_n -l)| \le |(y_n-x_n)| + |x_n-l| \le |(z_n - x_n)| + |x_n-l|=$$

$$|(z_n-l) + (l-x_n)| + |x_n-l| \le |z_n-l| + |x_n-l| + |x_n-l|$$.

Now we can "trap " the $$z_n, x_n$$ close to $$l$$.

Now $$\frac \epsilon 3 > 0$$ so there are $$N_1$$, and $$N_2$$ so that if $$n> \max(N_1, N_2)$$the we have $$|z_n -l| < \frac \epsilon 3$$ and $$|x_n -l|<\frac \epsilon 3$$.

SO $$|y_n -l| \le |z_n-l| + |x_n-l| + |x_n-l|<3\frac \epsilon 3 = \epsilon$$.

Let $$\epsilon>0$$ As $$x_n\rightarrow l$$ then you can find $$N_1 \in \mathbb{N}$$ such that whenever $$n\geq N_1$$ we have $$|x_n-l|<\epsilon$$ Similarly, $$\exists N_2 \in \mathbb{N}$$ such that whenever $$n\geq N_2$$ then $$|z_n-l|< \epsilon$$.

Therefore, since $$x_n\leq y_n \leq z_n$$ for each $$n\in \mathbb{N}$$ it follows that for $$N=$$ $$max(N_1,N_2)$$ we have $$N\geq N_1$$ and $$N\geq N_2$$. Therefore

$$-\epsilon+l$$ $$\leq$$ $$x_n$$ $$\leq y_n$$ $$\leq$$ $$z_n$$ $$\leq$$ $$\epsilon+ l$$ hence $$|y_n -l| \leq$$ $$\epsilon$$. As $$\epsilon>0$$ is arbitrary, $$y_n \rightarrow l$$.

Further note if $$a and $$b it is not necessarily the case that $$a-b<0$$. Consider the following counterexample $$5<10$$, $$1<10$$ but $$4<0$$ is clearly false. Further note that showing $$x_n = z_n$$ is not necessarily possible given the conditions of the hypothesis.

Further, "$$|x_n-l|<\epsilon$$ for some $$\epsilon>0$$ " does not follow from the definition. You need to be precise.