Introduction to Mathematical Analysis I - Second Edition

91 Let us now prove the converse. Suppose (1) and (2) are satisfied. Fix any ε > 0 and let δ > 0 satisfy (1) . Then g ( δ ) = sup x ∈ B 0 ( ¯ x ; δ ) ∩ D f ( x ) ≤ ` + ε . This implies limsup x → ¯ x f ( x ) = inf δ > 0 g ( δ ) ≤ ` + ε . Since ε is arbitrary, we get limsup x → ¯ x f ( x ) ≤ `. Again, let ε > 0. Given δ > 0, let x δ be as in (2) . Therefore, ` − ε < f ( x δ ) ≤ sup x ∈ B 0 ( ¯ x ; δ ) ∩ D f ( x ) = g ( δ ) . This implies ` − ε ≤ inf δ > 0 g ( δ ) = limsup x → ¯ x f ( x ) . It follows that ` ≤ limsup x → ¯ x f ( x ) . Therefore, ` = limsup x → ¯ x f ( x ) . Corollary 3.6.2 Suppose ` = limsup x → ¯ x f ( x ) . Then there exists a sequence { x k } in D such that { x k } converges to ¯ x , x k 6 = ¯ x for every k , and lim k → ∞ f ( x k ) = `. Moreover, if { y k } is a sequence in D that converges to ¯ x , y k 6 = ¯ x for every k , and lim k → ∞ f ( y k ) = ` 0 , then ` 0 ≤ ` . Proof: For each k ∈ N , take ε k = 1 k . By (1) of Theorem 3.6.1 , there exists δ k > 0 such that f ( x ) < ` + ε k for all x ∈ B 0 ( ¯ x ; δ k ) ∩ D . (3.11) Let δ 0 k = min { δ k , 1 k } . Then δ 0 k ≤ δ k and lim k → ∞ δ 0 k = 0. From (2) of Theorem 3.6.1 , there exists x k ∈ B 0 ( ¯ x ; δ 0 k ) ∩ D such that ` − ε k < f ( x k ) . Moreover, f ( x k ) < ` + ε k by ( 3.11 ) . Therefore, { x k } is a sequence that satisfies the conclusion of the corollary. Now let { y k } be a sequence in D that converges to ¯ x , y k 6 = ¯ x for every k , and lim k → ∞ f ( y k ) = ` 0 . For any ε > 0, let δ > 0 be as in (1) of Theorem 3.6.1 . Since y k ∈ B 0 ( ¯ x ; δ ) ∩ D when k is sufficiently large, we have f ( y k ) < ` + ε for such k . This implies ` 0 ≤ ` + ε . It follows that ` 0 ≤ ` .