314CHAPTER 7. IMPROPER INTEGRALS AND INDETERMINATE FORMS

7.4Improper Integrals

1.Suppose that f is continuous on (−∞, ∞) and g0(x) = f(x). Then define each of the following improper integrals:

c

(vii) lim (ean ) = elimn→∞ an

n→∞

(viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and

lim an = lim cn = L.

n→∞ n→∞

Then

lim bn = L.

n→∞

Proof. Suppose that {an}∞n=m converges to a and {bn}∞n=m converges to b. Let 1 > 0 be given. Then there exist natural numbers N and M such that

Part (i) Let > 0 be given and c =6 0. Let 1 = 2|c| and n ≥ N + M. Then by the inequalities (1) and (2), we get

|can − ca| = |c| |an − a|

<|c| 1

<.

=a + (−1)b

=a − b.

|anbn − ab| = |[(an − a) + a][(bn − b) + b] − ab|

= |(an − a)(bn − b) + (an − a)b + a(bn − b| ≤ |an − a| |bn − b| + |b| |an − a| + |a| |bn − b|

< 21 + |b| 1 + |a| 1

= 1( 1 + |b| + |a|)

≤ 1(1 + |b| + |a|)

≤ .

Part (v) First we assume that b > 0 and prove that

lim 1 = 1.

n→∞ bn b

318 CHAPTER 8. INFINITE SERIES

1 = (−a) −b

= ab .

This completes the proof of Part (v).

Part (vi) Since f(x) = xc is a continuous function,

c

Part (vii) Since f(x) = ex is a continuous function,

lim ean = elimn→∞ an = ea.

n→∞

Part (viii) Suppose that an ≤ bn ≤ cn for all n ≥ m and

lim an = L = lim cn = L.

n→∞ n→∞

Let > 0 be given. Then there exists natural numbers N and M such that

It follows that

lim bn = L.

n→∞

This completes the proof of this theorem.