Calculus: Analytic Geometry and Calculus, with Vectors

716 Appendix I

The first basic theorem shows us that there can be at most one number L

for which lim f (x) = L.

Theorem A

(2) If lim f(x) = L1, lim f(x) = L2 then L2 = L1.

x- a x-a

Let e be a given positive number. Then e/2 is also a positive number

which we could call el. The first hypothesis of the theorem implies that

there is a positive constant S1, such that

(3) if(x) - L11 <

whenever x 0 a and Ix - at < S1. The second hypothesis implies that

there is a positive number S2 such that

(4) If(x) - L21 <

whenever x 0 a and Ix - al < S2. Let S be the lesser of Si and S.

Then, when x 34 a and Ix - al < 5, the two inequalities (3) and (4)

both hold and hence

(5) IL2 - L11 = I [f(x) - L1 - [f(x) - L2}I

f(x)-Lll+lf(x)-L21 <2+2=e.

If we suppose that IL2 - L11 0 0, then we could let e be the positive

number JIL2 - L1I and reach the false conclusions that IL2 - LI1 <

J1L2 - L11 and 1 < - and 2 < 1. Therefore, IL2 - L11 = 0 and hence

L2 = L1. This proves Theorem A. The last part of the proof involves

a principle that is very often used. If h is a number and if Ihj < e when-

ever e > 0, then It = 0.

Theorem B

If b isa constant, then

(6) limb = b.

x- a

This theorem tells us that if,f(x) = b, where b is a constant, then

(7) lim f (x) = b.

x-a

To prove the result, let e > 0. Since f(x) - b = 0 for each x, we can

let 3 = e and conclude that I f(x) - bI < e when x 0 a and Ix - at < S.