6.7. THE LIMIT OF A SEQUENCE 105

(e) The limit as (x,y)→ (1,2) of the expression

−2yx2 +8yx+34y+3y3−18y2 +6x2−13x−20− xy2− x3

−y2 +4y−5− x2 +2x.

Hint: It might help to write this in terms of the variables (s, t) = (x−1,y−2) .

2. In the definition of limit, why must x be a limit point of D(f)? Hint: If x were not alimit point of D(f), show there exists δ > 0 such that B(x,δ ) contains no points ofD(f) other than possibly x itself. Argue that 33.3 is a limit and that so is 22 and 7and 11. In other words the concept is totally worthless.

6.7 The Limit Of A SequenceAs in the case of real numbers, one can consider the limit of a sequence of points in Fp.

Definition 6.7.1 A sequence {an}∞

n=1 converges to a, and write

limn→∞

an = a or an→ a

if and only if for every ε > 0 there exists nε such that whenever n≥ nε ,

|an−a|< ε.

In words the definition says that given any measure of closeness, ε, the terms of thesequence are eventually all this close to a. There is absolutely no difference between thisand the definition for sequences of numbers other than here bold face is used to indicate anand a are points in Fp.

Theorem 6.7.2 If limn→∞ an = a and limn→∞ an = a1 then a1 = a.

Proof: Suppose a1 ̸= a. Then let 0 < ε < |a1−a|/2 in the definition of the limit. Itfollows there exists nε such that if n ≥ nε , then |an−a| < ε and |an−a1| < ε. Therefore,for such n,

|a1−a| ≤ |a1−an|+ |an−a|< ε + ε < |a1−a|/2+ |a1−a|/2 = |a1−a| ,

a contradiction.As in the case of a vector valued function, it suffices to consider the components. This

is the content of the next theorem.

Theorem 6.7.3 Let an =(an

1, · · · ,anp)∈ Fp. Then limn→∞ an = a≡(a1, · · · ,ap) if and only

if for each k = 1, · · · , p,limn→∞

ank = ak. (6.7.8)