1549901369-Elements_of_Real_Analysis__Denlinger_

192 Chapter 4 11 Limits of Functions

Leto= min{o 1 ,o 2 }. Then, Vx E D(f + g),

0 < Ix - xol < o =? 0 < Ix - xol < 81 and 0 < Ix - xol <Oz

c c

=? lf(x) -LI <

2

and lg(x) - Ml< 2

c c

=? lf(x) -LI+ lg(x) -Ml < 2 + 2

=? lf(x) -L + g(x) - Ml < c, by 6 -inequality

=? l(f(x) + g(x)) - (L + M)I < c.

Therefore, lim (f(x) + g(x)) = L + M.
X--+Xo
( c) Exercise 5.
(d) Let c > 0.
Since lim f(x) = L, f is bounded on some deleted nbd. of xo; i.e., :J B > 0
X--+Xo
and :J 81 > 0 3 0 < Ix - xol < 81 =? lf(x)I :S B. Since lim f(x) = L, :J 82 >
x--+xo

0 3 Vx E D(f), 0 < Ix - xol < 02 =? lf(x) -LI <
2
(IM~ + l). Finally, since

x-+xo lim g(x) = M, :303 >^0 3 Vx E D(g),^0 <Ix -xol <^03 =? lg(x) -Ml< 2 cB.
Leto= min{o 1 , o 2 , 83 }. Then, Vx E D(f · g),

0 < Ix - xo I < o =? 0 < Ix - xo I < 81 and 82 and 03

c c

=? lf(x)I :SB, lf(x) - LI< 2 (IMI + l), and lg(x) -Ml< 2 B

c c

=? (IMI + l)lf(x) -LI< 2 and Blg(x) -Ml< 2

c c

=? IMllJ(x) -LI < 2 and lf(x)llg(x) -Ml< 2

c c

=? IMJ(x) - MLI < 2 and lf(x)g(x) - f(x)MI < 2

c c

=? lf(x)g(x) - f(x)MI < 2 and /f(x)M - LM/ < 2

c c

=? /f(x)g(x) - f(x)M + f(x)M -LMI <

2

+

2

by 6-inequality

=? /f(x)g(x) - LMI < c.