184 Chapter 4 • Limits of Functions
Proof. Apply the rules of quantifier negation^3 to Definition 4.1.1. •Theorem 4.1.8 (Uniqueness of Limits) A function cannot have more than
one limit as x --> xo.
Proof. Suppose lim J(x) =Land lim J(x) = M. We want to prove that
x-+xo x -+xo
L = M. We shall use the "forcing principle" [Theorem 1.5.9 (d)]. Let c: > 0.
c
Since lim f(x) = L, 361 > 0 3 \:Ix E V(J), 0 < lx-xol < 61 =;. IJ(x)-LI < -
2
.
X-+Xo
Since lim J(x) = M , 3 62 > 0 3 \:Ix E V(J), 0 < Ix - xol < 62 =::;. lf(x) - Ml <
X-+Xo
- 2
Let 6 = min{6 1 ,6 2 }. Then, \:Ix E V(J),
0 < Ix - xo I < 6 =::;. 0 < Ix - xo I < 61 and 0 < Ix - xo I < 62
c c
=::;. lf(x) - LI < 2 and IM - f~x)I; 2
=::;. IJ(x) - LI+ JM - f(x)I < 2 + 2
=::;. lf(x) - L + M - f(x)I :::; IJ(x) - LI+ IM - f(x)I < c
(by 6-inequality)
=::;. JM - LI < c.Thus, Ve: > 0, IM - LI < c:. By the forcing principle, L = M. •
The next theorem suggests that the extensive work with limits of sequences
in Chapter 2 will yield dividends in our study of limits of functions.Theorem 4.1.9 (Sequential Criterion for Limits of Functions)
lim J(x) = L ~\:/sequences {xn} in V(J) - {xo} 3 Xn--> Xo, J(xn)--> L.
X-+Xo
Proof. Part 1 (=::;.):Suppose that lim J(x) = L , and suppose that {xn}
X-+XQ
is a sequence in V(J) - {x 0 } 3 Xn--> x 0. To prove: J(xn) --> L. Let c > 0.
Since lim f(x) = L , 36 > 0 3 0 < Jx - xol < 6 =::;. lf(x) - LI< c:.
x-..xo
Since Xn --> xo and Xn -:/:-xo, 3 no E N 3 n :::'.: no =::;. 0 < lxn - xo I < 6.
Thus, n :::'.: no =:;> If (xn) - LI < c. That is, f (xn) --> L.
Therefore,\:/ sequences {xn} in V(J) - {xo} 3 Xn--> xo, J(xn)--> L.
Part 2 (<=):Suppose that\:/ sequences {xn} in V(J) - {xo} 3 Xn--> x 0 ,
f(xn) --> L. We want to prove that lim f(x) = L. For contradiction, s uppose
X-+Xo
it is not true that lim f(x) = L. By Lemma 4.1.7, this means that 3c > 0 3
x -+xo- See Appendix A.2 for rules governing quantifiers and their negations.