What are the uses of "squeezing"? Off hand, the uses of "squeezing" that I can think of are:


*

*showing that $\lim_{x\to0}\dfrac{\sin x}x = 1$, which is then used in finding derivatives (PS: I've just remembered this item showing $\tan'=\sec^2$ by squeezing, without first differentiating any other trigonometric functions nor finding any limits besides the one in the definition of differentiation.  I'm inclined to consider that another part of the same item on the list rather than a separate item, but I'm glad I also know this argument.);

*exercises like finding the limit of the above when $x\to0$ is changed to $x\to\infty$, or ascertaining the convergence or divergence of a series by doing something involving finding limits of that sort, etc.; where all of this is not subsequently used to derive other results;

*showing that there are functions that are differentiable everywhere but whose derivatives have discontinuities;

*Various things in probability, possibly the most prominent of which is the proof of the weak law of large numbers.  I'm adding this bullet point after writing the "PS:" below.

*an argument that I wrote.  Here are some specifics: Let
$$
N = \text{number of persons whose income strictly exceeds }x;
$$
$$
M = \text{total income of all whose income strictly exceeds }x.
$$
Doing a continuous approximation to discrete variables, we pretend these vary continuously as functions of $x$.  Although they may be non-one-to-one functions of $x$, they are easily seen to be one-to-one functions of each other.  I demonstrated this proposition:Lemma: Except when $x$ is within a closed interval on which $M$ and $N$ are constant as functions of $x$, we have $\dfrac{dM}{dN}=x$. This is readily shown by squeezing: If $\Delta x>0$ then $x<\dfrac{\Delta M}{\Delta N}\le x+\Delta x$ and if $\Delta x<0$ then $x+\Delta x<\dfrac{\Delta M}{\Delta N} \le x$.


Quite possibly there are other uses that I know of very well but that don't come to mind.  If you were to say to me "How do you prove $P$?" I might instantaneously know that it's by squeezing, but if I ask myself "What things are done by squeezing?", perhaps most of them don't come to mind.
So my question is: How shall we extend this bulleted list of applications of squeezing, listing items in order of their importance in the work of the generic working mathematician, including, but not limited to, uses in research, scholarship, pedagogy, and applications of mathematics to other fields?
PS: I'd forgotten this item, which might be what made me think of asking this question in the first place: Independent, random variables with equal distribution satisfy: $\lim_{n \to \infty}\mathbb{P}\left(X_{n+1} > \sum_{i = 1}^{n}X_i\right) = 0$
 A: Here's one tiny minor example:
If car goes sixty miles in two hours, its average speed during that time is $60/2=30$ miles per hour.  But what is its speed at an instant, when it goes $0$ miles in $0$ hours?  Conventionally one speaks of a limit of a difference quotient, but here is a definition by squeezing:
If car A overtakes car B at a particular instant and car B has constant speed (so that its speed at that instant is unproblematic), then car A may not be going fast than car B at that instant because car A may have been slowing down while approaching B, then exactly matching speeds at that instant, then speeding up.  But car A's speed at that instant was not slower than that of car B, the definition of whose speed at that instant is unproblematic.  If we can thus squeeze car A's speed between those of all cars at constant speeds whose speed is not greater than that of car A, and all such whose speed is not less than that of car A, then we define the instantaneous speed of car A.
This is the definition of "derivative" used in the book titled Calculus Unlimited.  (Books of that sort must fail in our present pedagogical culture because "we" (I refuse to take that offensive word literally here) deliberately push into taking calculus those students whom we know to be unable to properly learn calculus, let alone appreciate things like this.)
A: Any instance of the limit at some point of a sine or cosine function multiplied by a function that approaches zero at that point can be shown to be zero by squeezing. For example, $$\lim_{x \rightarrow0}x\cos x=0$$ because $$-x\leq x\cos x\leq x$$ and $$\lim_{x\rightarrow0}-x=\lim_{x\rightarrow0}x=0.$$ In addition, limits at plus or minus infinity can be calculated this way. As an example, $$\lim_{x\rightarrow-\infty}e^x\sin x=0.$$ It is bounded by the functions $e^x$ and $-e^x$, both of which approach $0$ as $x\rightarrow-\infty$.
Edit: This can be generalized to the case of any bounded function, not just the sine or cosine functions. For instance, $$\lim_{x\rightarrow0}\frac{x}{1+x^2}=0.$$ The bounded function in this instance is $f(x)=\frac{1}{1+x^2}$ and the function that approaches zero is $g(x)=x$.
Edit 2: All of this is, in fact, useless, unless one or both of the functions is discontinuous.
A: Here is an non-trivial application of squeezing which could (theoretically) be shown to first-semester calculus students, in the sense that it uses a bare minimum of machinery. 
It is well-known that if $f: \mathbb{R} \to \mathbb{R}$ is a continuous function that satisfies the exponential law $f(x+y) = f(x)f(y)$, then $f$ is in fact infinitely differentiable and satisfies $f'(x) = f'(0)f(x)$. [In fact, one can weaken the hypothesis, replacing continuity of $f$ by measurability or even local measurability, but let's keep this more at a calculus level.] The "hard part" is to show that $f'(0)$ exists; once we have that, it's an easy calculation. But how do we show the derivative exists? 
One famous proof goes like this: suppose we are in the nontrivial case where $f$ is not identically zero and therefore is everywhere positive. Then $\int_0^a f(t) dt$ is nonzero for some $a$ (the Riemann integral exists since $f$ is assumed continuous); let $C$ denote this quantity. Then 
$$Cf(x) = \int_0^a f(x)f(t)dt = \int_0^a f(x+t)dt = \int_x^{x+a} f(t)dt$$ 
where the right hand side is differentiable by the fundamental theorem of calculus. 
That's nice, but it does presuppose a certain amount of integration theory, i.e., machinery that we won't presume our first-semester students know. Similarly, we won't presume the students know the mean value theorem or the intermediate value theorem (or certainly not how to prove them!). All I would like to presume of our (ideal) students is that they know a set of reals with an upper/lower bound has a lub/glb. 
So, here is another method. First show that $f$ is a convex function, i.e., that for $x, y$ and $0 \leq t \leq 1$ we have $f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$. Sketch of proof: show it holds at $t = 1/2$ by invoking $a b \leq \frac{a^2 + b^2}{2}$ where $a = f(x/2)$ and $b = f(y/2)$. Iterate this to prove the inequality for all dyadic rationals $t \in [0, 1]$. Then extend to all $t \in [0, 1]$ by invoking continuity of $f$. 
It follows from convexity of $f$ that slopes of secant lines $\frac{f(t) - f(0)}{t}$ are increasing as a function of nonzero $t$. Thus, the slopes for positive $t$ have a lower bound given by $\frac{f(s)-f(0)}{s}$ for any $s < 0$, hence they have a greatest lower bound; this inf is the right-hand derivative 
$$R_0 = \lim_{t \to 0^+} \frac{f(t)-f(0)}{t}$$ 
since these secants decrease as $t$ decreases to $0$. By similar reasoning, the left-hand derivative 
$$L_0 = \lim_{t \to 0^-} \frac{f(t)-f(0)}{t}$$ 
exists, and we have $L_0 \leq R_0$. All we need to do is prove $L_0 = R_0$, or that $c := R_0/L_0 = 1$ (we know $1 \leq c$). 
Here is where we apply a squeezing argument. Define 
$$L_a := \lim_{h \to 0^-} \frac{f(a+h)-f(a)}{h}; \qquad R_a := \lim_{h\to 0^+} \frac{f(a+h)-f(a)}{h}.$$ 
One easily verifies that $R_a = f(a)R_0$ and $L_a = f(a)L_0$. Hence $R_a/L_a = c$ for any $a$. Also observe by our earlier reasoning that whenever $a \lt b$, we have 
$$R_a \lt \frac{f(b)-f(a)}{b-a} \lt L_b$$ 
so that $L_b/R_a \gt 1$. Telescoping, we have for every $n$ that 
$$\array{
\frac{L_1}{L_0} & = & (\frac{R_0}{L_0}\frac{L_{1/n}}{R_0})\cdot (\frac{R_{1/n}}{L_{1/n}}\frac{L_{2/n}}{R_{1/n}}) \cdot \ldots \cdot (\frac{R_{(n-1)/n}}{L_{(n-1)/n}} \frac{L_1}{R_{(n-1)/n}}) \\ 
 & \gt & \frac{R_0}{L_0} \frac{R_{1/n}}{L_{1/n}} \ldots \frac{R_{(n-1)/n}}{L_{(n-1)/n}} \\ 
 & = & c^n
}$$ 
so that $1 \leq c \lt (\frac{L_1}{L_0})^{1/n}$ for all $n \geq 1$. Thus $c = 1$ by a squeezing argument, and we are done. 
