squeeze rule 477

Comparing areas

1

–

2

x

–

2

1

–

2

1

–

2

x

–

2π

π

–

2

x

–

2π

1

–

2

sinx

–––

x

cosx– 1

––––––

x

π

–––

180

sinx

–––

x

x

–––

360

sinx

–––

x

x

–––

sinx

1

–––

cosx

attempted to show that πis not only irrational, but,

moreover, that it is a

TRANSCENDENTAL NUMBER

. This

would establish, once and for all, that √

–

πcannot possi-

bly be constructed. Unfortunately, Gregory did not suc-

ceed in this goal. It was not until a century later that

German mathematician J

OHANN

H

EINRICH

L

AMBERT

(1728–77) succeeded in proving that πis irrational,

and another century after that that C

ARL

L

OUIS

F

ERDI

-

NAND VON

L

INDEMANN

(1852–1939) finally proved in

1880 that πis indeed transcendental. As a consequence,

Lindemann had proved that the problem of squaring

the circle is unsolvable.

It is interesting to note that in 1914 S

RINIVASA

A

IYANGAR

R

AMANUJAN

(1887–1920) used a ruler-and-

compass construction akin to squaring a circle to find

the following remarkable approximation for πcorrect

to the ninth decimal place:

squeeze rule (sandwich result) This rule asserts that

if a function f(x) is sandwiched between two other

functions g(x) and h(x),at least for values xclose to a

number a(that is, we have g(x) ≤f(x) ≤h(x)), and if

limx→ag(x) = Land limx→ah(x) = L, then it must be the

case that limx→af(x) exists and equals Las well. For

example, the inequalities 1 – x≤f(x) ≤1 + x2imply

that limx→0f(x) = 1.

Perhaps the most important application of the

squeeze rule is the calculation of the limit , which

arises in computing the

DERIVATIVE

of the sine function.

In the diagram below we notice that the sector is sand-

wiched between two right triangles. Here the angle xis

given in

RADIAN MEASURE

. We have:

area of the small right triangle = · cos x· sin x

area of the sector = · π12=

(This is the fraction of the area of a full circle of

radius 1), and

area of the large right triangle · 1 · tan x.

From the picture, we see that · cosx· sinx≤≤

· tan x, which can be rewritten: .

Since limx→0cosx= 1 and limx→0= 1, it follows by

the squeeze rule that limx→0exists and equals 1.

This proves:

If xis measured in radians, then limx→0= 1.

(If xis measured in degrees, then the area of the sector is

given by · π. Chasing through the argument above

shows that, for xmeasured in degrees, limx→0=)

As an aside, the limit limx→0follows: