Introduction to Mathematical Analysis I - Second Edition

120 4.4 L’HOSPITAL’S RULE such x , since g 0 ( z ) 6 = 0 if x < z < α , Rolle’s theorem (Theorem 4.2.2 ) guarantees that g ( x ) 6 = g ( α ) . Therefore, for all x ∈ B + ( ¯ x ; δ 2 ) we can write, f ( x ) g ( x ) = f ( x ) − f ( α ) g ( x ) − g ( α ) 1 − g ( α ) g ( x ) 1 − f ( α ) f ( x ) . Now, define H α ( x ) = 1 − g ( α ) g ( x ) 1 − f ( α ) f ( x ) for x ∈ B + ( ¯ x ; δ 2 ) . Since lim x → ¯ x f ( x ) = lim x → ¯ x g ( x ) = ∞ , we have that lim x → ¯ x + H α ( x ) = 1. Thus, there exists a positive γ < δ 2 such that | H α ( x ) − 1 | < ε 2 K whenever x ∈ B + ( ¯ x ; γ ) . For any x ∈ B + ( ¯ x ; γ ) , applying Theorem 4.2.4 on the interval [ x , α ] , we can write [ f ( x ) − f ( α )] g 0 ( c ) = [ g ( x ) − g ( α )] f 0 ( c ) for some c ∈ ( x , α ) (note that, in particular, c ∈ B ( ¯ x : δ 1 ) ∩ ( a , b ) ). For such c we get f ( x ) g ( x ) = f 0 ( c ) g 0 ( c ) H α ( x ) . Since c ∈ B ( ¯ x : δ 1 ) ∩ ( a , b ) , applying ( 4.12 ) we get that, for x ∈ B + ( ¯ x ; γ ) = B + ( ¯ x ; γ ) ∩ ( a , b ) , f ( x ) g ( x ) − ` = f 0 ( c ) g 0 ( c ) H α ( x ) − ` = f 0 ( c ) g 0 ( c ) ( H α ( x ) − 1 )+ f 0 ( c ) g 0 ( c ) − ` ≤ f 0 ( c ) g 0 ( c ) | H α ( x ) − 1 | + f 0 ( c ) g 0 ( c ) − ` < K ε 2 K + ε 2 = ε . Setting δ 0 = γ completes the proof. Example 4.4.5 Consider the limit lim x → 0 ln x 2 1 + 1 3 √ x 2 . Here f ( x ) = ln x 2 , g ( x ) = 1 + 1 3 √ x 2 , ¯ x = 0, and we may take as ( a , b ) any open inteval containing 0. Clearly f and g satisfy the differentiability assumptions and g 0 ( x ) 6 = 0 for all x 6 = 0. Moreover, lim x → ¯ x f ( x ) = lim x → ¯ x g ( x ) = ∞ . We analyze the quotient of the derivatives. We have lim x → 0 2 / x − 2 3 1 3 √ x 5 = lim x → 0 − 3 3 √ x 5 x = lim x → 0 − 3 3 √ x 2 = 0 .