Squeeze theorem: sandwich to conclude

Statement

Squeeze theorem: if for all $n$ from some rank on:

$v_n \leq u_n \leq w_n$

and if $\lim v_n = \lim w_n = \ell$, then:

$\lim u_n = \ell$

Idea: $u_n$ is “trapped” between two sequences converging to the same limit. So it must go there too.

Function version

If $v(x) \leq f(x) \leq w(x)$ near $a$ (or $\pm\infty$), and $\lim v = \lim w = \ell$, then $\lim f = \ell$.

Application: sequences with cos/sin

$u_n = \dfrac{\cos(n)}{n}$. What limit?

$-1 \leq \cos(n) \leq 1$, so:

$-\dfrac{1}{n} \leq \dfrac{\cos(n)}{n} \leq \dfrac{1}{n}$

$-1/n \to 0$ and $1/n \to 0$, so $u_n \to 0$.

Corollary: limit $+\infty$

If $u_n \geq v_n$ and $v_n \to +\infty$, then $u_n \to +\infty$.

Same: if $u_n \leq v_n$ and $v_n \to -\infty$, then $u_n \to -\infty$.

Example: $u_n = n + \sin(n) \geq n - 1$. And $n - 1 \to +\infty$, so $u_n \to +\infty$.

Method

1. Bracket $u_n$ between two simpler expressions.
2. Compute limits of the brackets.
3. Check they’re equal (else fails).
4. Conclude by the theorem.

Traps

⚠️ If the two limits aren’t equal, we can not conclude. Find a tighter bracket.

⚠️ The bracket must hold for all $n$ (from some rank).

Common mistakes

1. Bracket too loose (limits of $v_n$ and $w_n$ differ).
2. Reversed inequalities.
3. Forgetting to prove the bracket (not just assert).

?À toi de jouer

u_n = sin(n)/n². Limit as n → ∞?

A.0B.1C.sin(∞)/∞ (indeterminate)D.does not exist

Valider