Introduction to Mathematical Analysis I - Second Edition

118 4.4 L’HOSPITAL’S RULE First we observe that the conditions of Theorem 4.4.1 hold. Here f ( x ) = 2 x + sin x , g ( x ) = x 2 + 3 x , and ¯ x = 0. We may take [ a , b ] = [ − 1 , 1 ] , for example, so that f and g are continuous on [ a , b ] and differentiable on ( a , b ) and, furthermore, f ( x ) g ( x ) is well defined on [ a , b ] \ { ¯ x } . Moreover, taking δ = 7 / 3, we get g 0 ( x ) = 2 x + 3 6 = 0 for x ∈ B ( ¯ x ; δ ) ∩ [ a , b ] . Finally we calculate the limit of the quotient of derivatives using Theorem 3.2.1 to get lim x → ¯ x f 0 ( x ) g 0 ( x ) = lim x → 0 2 + cos x 2 x + 3 = lim x → 0 2 + lim x → 0 cos x lim x → 0 2 x + 3 = 2 + 1 3 = 1 . It now follows from Theorem 4.4.1 that lim x → 0 2 x + sin x x 2 + 3 x = 1 as we wanted to show. Example 4.4.2 We will apply L’Hospital’s rule to determine the limit lim x → 1 3 x 3 − 2 x 2 + 4 x − 5 4 x 4 − 2 x − 2 . Here f ( x ) = 3 x 3 − 2 x 2 + 4 x − 5 and g ( x ) = 4 x 4 − 2 x − 2. Thus f ( 1 ) = g ( 1 ) = 0. Moreover, f 0 ( x ) = 9 x 2 − 4 x + 4 and g 0 ( x ) = 16 x 3 − 2. Since g 0 ( 1 ) = 14 6 = 0 and g 0 is continuous we have g 0 ( x ) 6 = 0 for x near 1. Now, lim x → 1 9 x 2 − 4 x + 4 16 x 3 − 2 = 9 14 . Thus, the desired limit is 9 14 as well. Example 4.4.3 If the derivatives of the functions f and g themselves satisfy the assumptions of Theorem 4.4.1 we may apply L’Hospital’s rule to determine first the limit of f 0 ( x ) / g 0 ( x ) and then apply the rule again to determine the original limit. Consider the limit lim x → 0 x 2 1 − cos x . Here f ( x ) = x 2 and g ( x ) = 1 − cos x so both functions and all its derivatives are continuous. Now g 0 ( x ) = sin x and, so, g 0 ( x ) 6 = 0 for x near zero, x 6 = 0. Also, f 0 ( 0 ) = 0 = g 0 ( 0 ) and g 00 ( x ) = cos x 6 = 0 for x near 0. Moreover, lim x → 0 f 00 ( x ) g 00 ( x ) = lim x → 0 2 cos x = 2 . By L’Hospital’s rule we get lim x → 0 f 0 ( x ) g 0 ( x ) = lim x → 0 f 00 ( x ) g 00 ( x ) = lim x → 0 2 cos x = 2 . Applying L’Hospital’s rule one more time we get lim x → 0 x 2 1 − cos x = lim x → 0 f ( x ) g ( x ) = lim x → 0 f 0 ( x ) g 0 ( x ) = 2 .