Limits Of Trigonometric Functions.pdf 6o332z

72

2

Limits and Continuity of Functions

2. 5

Limits of Trigonometric Functions Whenever we discuss limits of trigonometric expressions involving sin t, cos x, tan 0, etc., we shall assume that each variable represents a real number or the radian measure of an angle. The following result is important for future developments.

[

Theorem (2.18)

y

x

Jim sin t = 0 r-o

Proof Let us first prove that lim 1 _ 0 + sin t = 0. Since we are only interested in positive values of t near zero, there is no loss of generality in assuming that 0 < t < n/2. Let U be the circle of radius 1 with center at the origin of a rectangular coordinate system, and let A be the point (1, 0). If, as illustrated in Figure 2.17, P(x, y) is the point on U that corresponds to t, then by (1.26) the radian measure of angle AOP is t. Referring to the figure we see that O

or, since y = sin t, FIGURE 2.17

0 <sin t < t. Since lim 1 _ 0 + t = 0, it follows from the Sandwich Theorem (2.14) that lim 1_ 0 + sin t = 0. To complete the proof it is sufficient to show that lim 1 _ 0 _ sin t = 0. If - n/2 < t < 0, then 0 < - t < n/2 and hence, from the first part of the proof, 0 < sin ( - t) < - t. Multiplying the last inequality by -1 and using the fact that sin ( - t) = - sin t gives us t <sin t < 0.

Since lim 1_ 0 _ t = 0, it follows from the Sandwich Theorem that lim 1_ 0 _ sin t = 0. D

Corollary (2.19)

Jim cost= 1 r-o

Proof Since sin 2 t + cos 2 t = 1, it follows that cost = ±Jl - sin 2 t. If -n/2 < t < n/2, then cost is positive, and hence cost= J1 - sin 2 t. Consequently,

Jim cost= limJl - sin 2 t = J!im(l - sin 2 t) r-o t-o r-o = j1--=o = 1.

D

Limits of Trigonometric Functions

2.5

73

For our work in Section 3.4 it will be essential to know the limits of (sin t)/t and (l - cos t)/t as t approaches 0. These are established in Theorems (2.21) and (2.22). In the proof of (2.21) we shall make use of the following result.

Theorem (2.20)

If (J is the radian measure of a central angle of a circle of radius r, then the area A of the sector determined by 0 is A=

tr e. 2

Proof

A typical central angle 0 and the sector it determines are shown in Figure 2.18. The area of the sector is directly proportional to 0, that is, A= kO

for some real number k. For example, the area determined by an angle of 2 radians is twice the area determined by an angle of 1 radian. In particular, if (J = 2n, then the sector is the entire circle, and A = nri. Thus

FIGURE 2.18

or

nri = k(2n),

k = !ri

and therefore D

. sin t lim--= 1

Theorem (2.21)

1-+0

t

Proof If 0 < t < n/2 we have the situation illustrated in Figure 2.19 where U is a unit circle. If A 1 is the area of triangle AOP, Ai the area of circular sector AOP, and A3 the area of triangle AOQ, then y

Using the formula for the area of a triangle and Theorem (2.20) we obtain 0

M(x, 0)

x

A 1 = !(l)d(M, P) =

Ai = !(1) 2 t =

u Thus

=

t sin t

!t

A3 = !(l)d(A, Q) = FIGURE 2.19

ty

t tan t.

t sin t < !t < t tan t.

74

2 Limits and Continuity of Functions Dividing by t sin t and using the fact that tan t = sin t/cos t gives us t

1 cost

<--<-sin t

or, equivalently, sin t I> - - >cost. t

If - n/2 <

t

< 0, then 0 < I >

t

< n/2, and from the result just established,

sin(-t) > cos ( -t

t ).

Since sin ( -t) = -sin t and cos ( - t) = cost, this inequality reduces to ( *). This shows that ( *) is also true if - n/2 < t < 0, and hence is true for every tin the open interval ( -n/2, n/2) except t = 0. Since lim1_ 0 cost = 1, and (sin t)/t is always between cost and 1, it follows from the Sandwich Theorem that t 1I. msin - - = 1.

r-o

D

t

Roughly speaking, Theorem (2.21) implies that if t is close to 0, then (sin t)/t is close to 1. Another way of stating this is to write sin t ::::::: t for small values of t. It is important to that if t denotes an angle, then radian measure must be used in Theorem (2.21) and in the approximation formula sin t ::::::: t. To illustrate, trigonometric tables or a calculator show that to five decimal places, sin (0.06) : : : : 0.05996 sin (0.05) : : : : 0.04998 sin (0.04) : : : : 0.03999 sin (0.03) : : : : 0.03000.

. 1 - cost 0 1Im =

Theorem (2.22)

t

Proof

We may change the form of (1 - cos t)/t as follows:

1 - cos t

1 - cos t 1 + cos t 1 +cost 1 - cos 2 t t(l + cost) sin 2 t t(l + cost) sin t

sin t

1 + cost.

Limits of Trigonometric Functions

2 .5

75

Consequently, sin t ) . 1 - cos t 1. (sin t IIm =Im - - · - - - r-o

r-o

t

t

t) (i·Im---sin t )

. sin = (I Im-1_0 t =

Example I

1 + cost

r-o 1 + cost

1. (i ~ 1) = 1.-o = 0.

D

Find limx-o (sin 5x)/2x.

Solution We cannot apply Theorem (2.21) directly, since the given expression is not in the form (sin t)/t. However, we may introduce this form (with t = 5x) by using the following algebraic manipulation:

. sin 5x IIm--= x-o 2x

. 1 sin 5x 11m--x-o 2 x

=Jim~ sin 5x x-o 2 5x = ~ Jim sin 5x . 2 x-o 5x It follows from the definition of limit that x --+ 0 may be replaced by 5x Hence, by Theorem (2.21), with t = 5x, we see that

--+

0.

Jim sin 5x = ~ (1) = ~ . x-o 2x 2 2 Warning: When working problems of this type, that sin 5x =I= 5 sin x. •

Example 2 Find lim,_ 0 (tan t)/2t. Solution

Using the fact that tan t = sin t/cos t, Jim tan t 2t

1-0

=Jim(~. sin t.

_1_)

r-o 2 t cos t = t. 1 . 1 = t.

•

Example 3 Find limx-o (2x + 1 - cos x)/3x. Solution We plan to use Theorem (2.22). With this in mind we begin by isolating the part of the quotient that involves (1 - cos x)/x and then proceed as follows.

. 2x + 1 - cos x . (2x 1 - cos x) hm = 1Im - + - - - x-o 3x x-o 3x 3x = Jim (2x) + Jim ~ (1 - cos x) x-o 3x x-o 3 x . 2 1 . 1 - cos x = IIm-+- 1I m - - - x-o 3 3 x-o x =1+t·0=1

•

76

2 Limits and Continuity of Functions

2.5

Exercises

Find the limits in Exercises 1-26. . x hm-.x-o sm x

x-o

sin 3 t

2 +sin x x-o 3+x

6 lim

13

x-o

21

.

sin (x/2)

x-0

x

18 h m - - -

sin t x +tan x

sin x

sin 2 2t

20 lim-2-

:243

x2 + 1

1-cosx x213

cost

1_ 0 I - sin t

10 12

lim

1 - 2x

x-o

14 Jim x-o

16 Jim

x-o cot x

Jim a 2 csc 2 a

2 cos x + cos x xz 2

24

•-0 cos (v + n/2) v

v-o

x sin x lim-x-o x2 + 1 2 -

csc 2x

22 Jim--

lim x cot x

25 Jim

26

rI m sin 3x --

x-o sin 5x

sin 2 (x/2) x-o sin x lim

Establish the limits in Exercises 27-30, where a and bare any nonzero real numbers. sin ax

a

bx

b

sin ax

a

x-o sin bx

b

27 lim--=x-o

x cos x - x 2 2x

1

1-0

x-o

1 - cos 31

x-o x +cos x

41 2 + 3t sin t lim t2 1-0

15 lim

8 lim

38

sin (-3x) 4x x-o lim

19 lim

1-0

2 cos 0 - 2

9 Jim

x-o

.y;._

38 +sin 8 8 o-o

5 Jim

11

1-0

1 - cost

4 lim

1-0 (21)3

o-o

sin x

2 lim--

3 lim--

7 lim

17 lim

29 lim--=-

sin t

28 Jim

1 - cos ax

x-o

30

bx

=0

I' cos ax 1m--=l

x-o cos bx

1_ 0 1 +cost

2. 6

Continuous Functions In arriving at the definition of limx-af(x) we emphasized the restriction x =I a. Several examples in preceding sections have brought out the fact that limx-a f (x) may exist even though f is undefined at a. Let us now turn our attention to the case in which a is in the domain off If f is defined at a and limx-a f(x) exists, then this limit may, or may not, equalf(a). Iflimx-a f(x) = f(a) then f is said to be continuous at a according to the next definition.

Definition (2.23)

A function f is continuous at a number a if the following three conditions are satisfied. (i)

f

(ii)

Jim f (x) exists.

(iii)

Jim f(x) = f(a).

is defined on an open interval containing a.

If f is not continuous at a, then we say it is discontinuous at a, or has a discontinuity at a.