Calculus: Analytic Geometry and Calculus, with Vectors

718

Let a and ei and e2 be positive numbers.
S1 and S2 such that

Appendix 1

Choose positive numbers

(17) I f(x) - LI < ei (0 < Ix - aI < S1),

(18) Ig(x) - MI < e2 (0 < Ix - al < S2).

Let S be the lesser of S1 and S2. Then, when 0 < Ix - al < S, we have

(19) I [f(x) + g(x)1 - [L + M1I = I[f(x) - L] + [g(x) - MII

< If(x) -LI +Ig(x) - MI <el+f2.

If we set El = f2 = e/2, then (16) and (19) give

(20) I [f(x) + g(x)l - [lim f(x) + lim g(x)lI < Ex-.a x-+a

when 0 < Ix - at < S. This proves (13).

To prove (14), we bridge the gap between f(x)g(x) and LM by sub-

tracting and adding the term f(x)M and using (17) and (18) to obtain

(21) if(x)g(x) - LMI = I [f(x)g(x) - f(x)Ml + [f(x)M - LM]I

If(x)I Ig(x) - MI + If(x) - LI IMI

< (ILI + El)E2 + e1M

when 0 < Ix - at < S. If we choose el such that e1IMI < E/2 and after-

ward choose E2 such that (ILI + E1)e2 < E/2, then (21) gives

(22) If(x)g(x) - LMI < E (0 < Ix - at < S).

This proves (14). To prove (15), we begin by proving that

(23) 1a

g(x) M

when lim g(x) = M and M s 0. The more general result (15) will then

r-. a

follow from (14) and the fact that

(24) lm lm [ f (x)1 1 = urnI 1[urnI ].

,a g(x) [g(x)l x-.a x-'a g(x)

To prove (23), we suppose that M 0, that e2 has been chosen such that

0 < E2 < IMI/2, and that S has been chosen such that

(25) Ig(x) - MI < E2

whenever 0 < Ix - at < S. Then

(26) IMI = Ig(x) - M - g(x)I s Ig(x) - MI + Ig(x)I < e2 + Ig(x)I

and hence

(27) I8(x)I>IMI-E2>M-M=M