Introduction to Mathematical Analysis I - Second Edition

117 4.3.5 B Let f be twice differentiable on an open interval I . Suppose that there exist a , b , c ∈ I with a < b < c such that f ( a ) < f ( b ) and f ( b ) > f ( c ) . Prove that there exists d ∈ ( a , c ) such that f 00 ( d ) < 0. 4.3.6 B Prove that the function f defined in Exercise 4.1.11 is not monotone on any open interval containing 0. 4.4 L’HOSPITAL’S RULE We now prove a result that allows us to compute various limits by calculating a related limit involving derivatives. All four theorems in this section are known as l’Hospital’s Rule . For this section, we assume a , b ∈ R with a < b . Theorem 4.4.1 Suppose f and g are continuous on [ a , b ] and differentiable on ( a , b ) . Suppose f ( ¯ x ) = g ( ¯ x ) = 0, where ¯ x ∈ [ a , b ] . Suppose further that there exists δ > 0 such that g 0 ( x ) 6 = 0 for all x ∈ B ( ¯ x ; δ ) ∩ [ a , b ] , x 6 = ¯ x . If lim x → ¯ x f 0 ( x ) g 0 ( x ) = `, then lim x → ¯ x f ( x ) g ( x ) = `. (4.9) Proof: Let { x k } be a sequence in [ a , b ] that converges to ¯ x and such that x k 6 = ¯ x for every k . By Theorem 4.2.4 , for each k , there exists a sequence { c k } , with c k between x k and ¯ x , such that [ f ( x k ) − f ( ¯ x )] g 0 ( c k ) = [ g ( x k ) − g ( ¯ x )] f 0 ( c k ) . Since f ( ¯ x ) = g ( ¯ x ) = 0, and g 0 ( c k ) 6 = 0 for sufficiently large k , we have f ( x k ) g ( x k ) = f 0 ( c k ) g 0 ( c k ) . Under the assumptions that g 0 ( x ) 6 = 0 for x near ¯ x and g ( ¯ x ) = 0, we also have g ( x k ) 6 = 0 for sufficiently large k . By the squeeze theorem (Theorem 2.1.6 ) , { c k } converges to ¯ x . Thus, lim k → ∞ f ( x k ) g ( x k ) = lim k → ∞ f 0 ( c k ) g 0 ( c k ) = lim x → ¯ x f 0 ( x ) g 0 ( x ) = `. Therefore, ( 4.9 ) follows from Theorem 3.1.2 . Example 4.4.1 We will use Theorem 4.4.1 to show that lim x → 0 2 x + sin x x 2 + 3 x = 1 .