The funnelling theorem:

Consider the picture below. I think we all would agree that as x approaches a that both f(x) and h(x) approach b. Now suppose that we have a third function g which satisfies $h(x) \leq g(x) \leq f(x)$ for all x in the domain of f, g and h (which we are taking to be the same set). The only possibility is for g(x) to approach b as well. This observation goes by many names: The Pinching Theorem, The Squeeze Theorem, The (Ham) Sandwich Theorem. I prefer to call it The Funnel Theorem as the values of g are being funnelled between the values of f and h toward the common value b.

Here is a nice application of this idea to the problem of analyzing the behaviour of $\sin(h)/h$ as h approaches , where numerical evidence suggests that $\sin(h)/h$ approaches 1 as h approaches .

From the diagram below, if $h\in(0,\pi/2)$ we see by comparing the areas of the two triangles with the area of the sector that

$$\frac{1}{2}\sin(h) \leq \frac{1}{2}h \leq \frac{1}{2}\frac{\sin(h)}{\cos{h}}$$

Since $\sin(h) \gt 0$ for the values of h we have in mind, dividing through by $\sin(h)/2$ yields

$$1 \leq \frac{h}{\sin(h)} \leq \frac{1}{\cos{h}}$$

Now taking reciprocals of all the quantities yields

$$\cos(h) \leq \frac{\sin(h)}{h} \leq 1.$$

If we believe that $\cos(h)$ approaches 1 as h approaches , then The Funnelling Theorem tells us that $\sin(h)/h$ approaches 1 as h approaches through positive values. However since $\cos(-h) = \cos(h)$ and $\sin(-h)/(-h) = \sin(h)/h$, our inequality is actually valid for $h\in(-\pi/2),0)$, and $\sin(h)/h$ approaches 1 as h approaches through negative values and positive values.

Compare this graph with one from Lecture 2.