4.4. THE LIMIT OF A SEQUENCE 61

≤ 2|b|∥an−a∥+ 2∥a∥

|b|2|∥bn−b∥| .

Now choose n2 so large that if n≥ n2, then

∥an−a∥< ε |b|4

, and |bn−b|< ε |b|2

4(∥a∥+1).

Letting nε > max(n1,n2) , it follows that for n≥ nε ,∥∥∥∥an

bn− a

b

∥∥∥∥≤ 2|b|∥an−a∥+ 2∥a∥

|b|2|bn−b|< 2

|b|ε |b|

4+

2∥a∥|b|2

ε |b|2

4(∥a∥+1)< ε.

Another very useful theorem for finding limits is the squeezing theorem.

Theorem 4.4.9 In case an,bn ∈ R, suppose limn→∞ an = a = limn→∞ bn and an ≤cn ≤ bn for all n large enough. Then limn→∞ cn = a.

Proof: Let ε > 0 be given and let n1 be large enough that if n≥ n1, |an−a|< ε/2 and|bn−a|< ε/2. Then for such n,|cn−a| ≤ |an−a|+ |bn−a|< ε. The reason for this is thatif cn ≥ a, then |cn−a|= cn−a≤ bn−a≤ |an−a|+ |bn−a| because bn ≥ cn. On the otherhand, if cn ≤ a, then

|cn−a|= a− cn ≤ a−an ≤ |a−an|+ |b−bn| .

As an example, consider the following.

Example 4.4.10 Let cn ≡ (−1)n 1n and let bn =

1n , and an =− 1

n . Then you may easily showthat limn→∞ an = limn→∞ bn = 0. Since an ≤ cn ≤ bn, it follows limn→∞ cn = 0 also.

Theorem 4.4.11 limn→∞ rn = 0. Whenever |r|< 1. Here r ∈ F.

Proof: If 0 < r < 1 if follows r−1 > 1. Why? Letting α = 1r − 1, it follows r = 1

1+α.

Therefore, by the binomial theorem, 0 < rn = 1(1+α)n ≤ 1

1+αn . Therefore, limn→∞ rn = 0 if0 < r < 1. In general, if |r|< 1, |rn|= |r|n→ 0 by the first part.

An important theorem is the one which states that if a sequence converges, so doesevery subsequence. You should review Definition 4.2.4 on Page 56 at this point.

Theorem 4.4.12 Let {xn} be a sequence with limn→∞ xn = x and let{

xnk

}be a

subsequence. Then limk→∞ xnk = x.

Proof: Let ε > 0 be given. Then there exists nε such that if n > nε , then ∥xn− x∥< ε.Suppose k > nε . Then nk ≥ k > nε and so

∥∥xnk − x∥∥< ε showing limk→∞ xnk = x as claimed.

Theorem 4.4.13 Let {xn} be a sequence of real numbers and suppose each xn ≤ l(≥ l) for all n large enough, and limn→∞ xn = x. Then x≤ l (≥ l) . More generally, suppose{xn} and {yn} are two sequences of real numbers such that limn→∞ xn = x and limn→∞ yn =y. Then if xn ≤ yn for all n sufficiently large, then x≤ y.

4.4. THE LIMIT OF A SEQUENCE 61< — |la, —al| + b, —b|\|.Now choose nz so large that if n > nz, then€|b| e|b|*an —a|| < ——, and |b, —b| < ———_..Letting ne > max (71,72), it follows that for n > ne,2 Ilalo?22 eb 2IKIal|__ ed|b] 4 |p|? 4(\lal| +1)an_fby2S pp lan all + |b, —b| < <€E. IAnother very useful theorem for finding limits is the squeezing theorem.Theorem 4.4.9 in case An, by € R, suppose limy4.0 An = A = Mp0 Dy and ay <Cn < by for all n large enough. Then limp. Cn = a.Proof: Let € > 0 be given and let n; be large enough that if n > m1, |a, —a| < €/2 and|b, —a| < €/2. Then for such n,|c, — a] < |ay — a|+|b, —a| < €. The reason for this is thatif Cy > a, then |cy — al = cy —a < by —a < lay — a| +|by — al because by, > cy. On the otherhand, if c, <a, thenlCn —@| =a—Cy Sa—ay <|a—a,|+|b—b,|.As an example, consider the following.Example 4.4.10 Let c, =(—1)" i and let by, = i, and ay = —i. Then you may easily showthat limp—soo Ay = iMyy00 by = 0. Since dyn < Cn < Dy, it follows limy 4.0 Cn = 0 also.Theorem 4.4.11 lim,_,.. 7° =0. Whenever |r| <1. Here r € F.Proof: If 0 <r <1 if follows r-! > 1. Why? Letting @ = 1 —1, it follows r= +r 1+a@°Therefore, by the binomial theorem, 0 < r’ = may < ee Therefore, lim,_,..r” = 0 if0<r< 1. In general, if |r| <1, |r”| = |r|” — 0 by the first part. §JAn important theorem is the one which states that if a sequence converges, so doesevery subsequence. You should review Definition 4.2.4 on Page 56 at this point.Theorem 4.4.12 Ler {xn} be a sequence with limy 0X, = x and let {Xn } beasubsequence. Then limy-5..Xn, = X.Proof: Let € > 0 be given. Then there exists ng such that if n > ne, then ||x, —x|| <e.Suppose k > ng. Then ny > k > ng and so || xn, — x|| < € showing limy_,.. X,, =x as claimed.Theorem 4.4.13 Lez {xn} be a sequence of real numbers and suppose each x, <1(> 1) for all n large enough, and limy-400Xpn = x. Then x <1(> 1). More generally, suppose{xn} and {yn} are two sequences of real numbers such that limy—y0.Xp = xX and liMy 500 Yn =y. Then if x <n for all n sufficiently large, then x < y.