Introduction to Mathematical Analysis I - Second Edition

119 Example 4.4.4 Let g ( x ) = x + 3 x 2 and let f : R → R be given by f ( x ) = ( x 2 sin 1 x , if x 6 = 0; 0 , if x = 0 . Now consider the limit lim x → 0 f ( x ) g ( x ) = lim x → 0 x 2 sin 1 x x + 3 x 2 . Using the derivative rules at x 6 = 0 and the definition of derivative at x = 0 we can see that f is differentiable and f 0 ( x ) = ( 2 x sin 1 x − cos 1 x , if x 6 = 0; 0 , if x = 0 , However, f 0 is not continuous at 0 (since lim x → 0 f 0 ( x ) does not exist) and, hence, L’Hospital’s rule cannot be applied in this case. On the other hand lim x → 0 x 2 sin 1 x x + 3 x 2 does exist as we can see from lim x → 0 x 2 sin 1 x x + 3 x 2 = lim x → 0 x sin 1 x 1 + 3 x = lim x → 0 x sin 1 x lim x → 0 ( 1 + 3 x ) = 0 . Theorem 4.4.2 Let a , b ∈ R , a < b , and ¯ x ∈ ( a , b ) . Suppose f , g : ( a , b ) \{ ¯ x } → R are differentiable on ( a , b ) \ { ¯ x } and assume lim x → ¯ x f ( x ) = lim x → ¯ x g ( x ) = ∞ . 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 ` ∈ R and lim x → ¯ x f 0 ( x ) g 0 ( x ) = `, (4.10) then lim x → ¯ x f ( x ) g ( x ) = `. (4.11) Proof: Since lim x → ¯ x f ( x ) = lim x → ¯ x g ( x ) = ∞ , choosing a smaller positive δ if necessary, we can assume that f ( x ) 6 = 0 and g ( x ) 6 = 0 for all x ∈ B ( ¯ x ; δ ) ∩ ( a , b ) . We will show that lim x → ¯ x + f ( x ) g ( x ) = ` . The proof that lim x → ¯ x − f ( x ) g ( x ) = ` is completely analogous. Fix any ε > 0. We need to find δ 0 > 0 such that | f ( x ) / g ( x ) − ` | < ε whenever x ∈ B + ( ¯ x ; δ 0 ) ∩ ( a , b ) . From ( 4.10 ) , one can choose K > 0 and a positive δ 1 < δ such that f 0 ( x ) g 0 ( x ) ≤ K and f 0 ( x ) g 0 ( x ) − ` < ε 2 (4.12) whenever x ∈ B ( ¯ x ; δ 1 ) ∩ ( a , b ) , x 6 = ¯ x . Fix α ∈ B + ( ¯ x ; δ 1 ) ∩ ( a , b ) (in particular, α > ¯ x ). Since lim x → ¯ x f ( x ) = ∞ , we can find δ 2 > 0 such that δ 2 < min { δ 1 , α − ¯ x } and f ( x ) 6 = f ( α ) for x ∈ B + ( ¯ x ; δ 2 ) ∩ ( a , b ) = B + ( ¯ x ; δ 2 ) . Moreover, for