1549901369-Elements_of_Real_Analysis__Denlinger_

4.2 Algebra of Limits of Functions 193

Therefore, lim (f(x)g(x)) = LM.

X-+Xo

(e) Let E. > 0. Since lim g(x) = M "I-0, g is bounded away from O on some

X-+Xo

deleted nbd. of xo. In fact, by Theorem 4.2.9, 381 > 0 3 x EN~, (x 0 ) n D(g) *

lg(x)I >^1 ~^1. Also, si nce lim g(x) = M, 3 82 > 0 3 Vx E D(g), 0 < Ix - xol <

X--tXQ

82 * lg(x) - Ml< e.1~1

2

. Leto= min{8 1 , 82}. Then, Vx ED(~),

0 < Ix - Xo I < o * 0 < Ix - xo I < 81 and 0 < Ix - xo I < 82

• lg(x)I > l~I

1 2

* lg(x)I < IMI

and

and

e.IMl2

lg(x) - Ml < -

2

e.IMl2

lg(x) - Ml< -

2

lg(x) - Ml 1 lg(x) - Ml 2 e.IMl^2

* lg(x)llMI = lg(x)I. IMI < IMI. 2IMI = E.

1

1 11 1M-g(x)1 lg(x)-MI

* g(x) - M = g(x)M = lg(x)llMI < e.

Therefore, lim (-(

1

X->XQ g X )) =Ml.

(f) Exercise 8.

(g) We postpone the proof of (g) until we have discussed an alternate
method of proof, which follows. •

Alternate proof of Theorem 4 .2.11 using the "sequential crite-
rion." The sequential criterion for limits of functions (Theorem 4.1.9) pro-
vides a very powerful technique for proving theorems about limits of functions.
It enables us to use the power of the theory of sequences developed in Chapter

2. As examples, we use it to gi ve alternate (much easier) proofs of Theorem
   4.2.11 Parts (d) and (g).

Theorem 4.2.11 (d): If lim f(x) = L and lim g(x) = M, then
x-+xo x-+xo
lim (f(x)g(x)) = LM. (Assuming xo is a cluster point of D(f) n D(g).)
X-+Xo