Introduction to Mathematical Analysis I - Second Edition

66 3.1 LIMITS OF FUNCTIONS Proof: Choose ε > 0 such that ` 1 + ε < ` 2 − ε (equivalently, such that ε < ` 2 − ` 1 2 ). Then there exists δ > 0 such that ` 1 − ε < f ( x ) < ` 1 + ε and ` 2 − ε < g ( x ) < ` 2 + ε for all x ∈ B ( ¯ x ; δ ) ∩ D , x 6 = ¯ x . Thus, f ( x ) < ` 1 + ε < ` 2 − ε < g ( x ) for all x ∈ B ( ¯ x ; δ ) ∩ D , x 6 = ¯ x . The proof is now complete. Theorem 3.1.7 Let f , g , h : D → R and let ¯ x be a limit point of D . Suppose there exists δ > 0 such that f ( x ) ≤ g ( x ) ≤ h ( x ) for all x ∈ B ( ¯ x ; δ ) ∩ D , x 6 = ¯ x . If lim x → ¯ x f ( x ) = lim x → ¯ x h ( x ) = ` , then lim x → ¯ x g ( x ) = `. Proof: The proof is straightforward using Theorem 2.1.6 and Theorem 3.1.2 . Remark 3.1.8 We will adopt the following convention. When we write lim x → ¯ x f ( x ) without speci- fying the domain D of f we will assume that D is the largest subset of R such that if x ∈ D , then f ( x ) results in a real number. For example, in lim x → 2 1 x + 3 we assume D = R \{− 3 } and in lim x → 1 √ x we assume D = [ 0 , ∞ ) . Exercises 3.1.1 Use the definition of limit to prove that (a) lim x → 2 3 x − 7 = − 1. (b) lim x → 3 ( x 2 + 1 ) = 10. (c) lim x → 1 x + 3 x + 1 = 2. (d) lim x → 0 √ x = 0. (e) lim x → 2 x 3 = 8. 3.1.2 Prove that the following limits do not exist. (a) lim x → 0 x | x | . (b) lim x → 0 cos ( 1 / x ) . 3.1.3 Let f : D → R and let ¯ x be a limit point of D . Prove that if lim x → ¯ x f ( x ) = ` , then lim x → ¯ x | f ( x ) | = | ` | . Give an example to show that the converse is not true in general.