Introduction to Mathematical Analysis I - Second Edition

68 3.2 LIMIT THEOREMS Proof: Let us first prove (a) . Let { x n } be a sequence in D that converges to ¯ x and x n 6 = ¯ x for every n . By Theorem 3.1.2 , lim n → ∞ f ( x n ) = ` and lim n → ∞ g ( x n ) = m . It follows from Theorem 2.2.1 that lim n → ∞ ( f ( x n )+ g ( x n )) = ` + m . Applying Theorem 3.1.2 again, we get lim x → ¯ x ( f + g )( x ) = ` + m . The proofs of (b) and (c) are similar. Let us now show that if m 6 = 0, then ¯ x is a limit point of e D . Since ¯ x is a limit point of D , there is a sequence { u k } in D converging to ¯ x such that u k 6 = ¯ x for every k . Since m 6 = 0, it follows from an easy application of Theorem 3.1.6 that there exists δ > 0 with g ( x ) 6 = 0 whenever 0 < | x − ¯ x | < δ , x ∈ D . This implies x ∈ e D whenever 0 < | x − ¯ x | < δ , x ∈ D . Then u k ∈ e D for all k sufficiently large, and hence ¯ x is a limit point of e D . The rest of the proof of (d) can be completed easily following the proof of (a) . Example 3.2.1 Consider f : R \ {− 7 } → R given by f ( x ) = x 2 + 2 x − 3 x + 7 . Then, combining all parts of Theorem 3.2.1 , we get lim x →− 2 f ( x ) = lim x →− 2 ( x 2 + 2 x − 3 ) lim x →− 2 ( x + 7 ) = lim x →− 2 x 2 + lim x →− 2 2 x − lim x →− 2 3 lim x →− 2 x + lim x →− 2 7 = ( lim x →− 2 x ) 2 + 2 lim x →− 2 x − lim x →− 2 3 lim x →− 2 x + lim x →− 2 7 = ( − 2 ) 2 + 2 ( − 2 ) − 3 − 2 + 7 = − 3 5 . Example 3.2.2 We proceed in the same way to compute the following limit. lim x → 0 1 +( 2 x − 1 ) 2 x 2 + 7 = lim x → 0 1 + lim x → 0 ( 2 x − 1 ) 2 lim x → 0 x 2 + lim x → 0 7 = 1 + 1 0 + 7 = 2 7 . Example 3.2.3 We now consider lim x →− 1 x 2 + 6 x + 5 x + 1 . Since the limit of the denominator is 0 we cannot apply directly part (d) of Theorem 3.2.1 . Instead, we first simplify the expression keeping in mind that in the definition of limit we never need to evaluate the expression at the limit point itself. In this case, this means we may assume that x 6 = − 1. For any such x we have x 2 + 6 x + 5 x + 1 = ( x + 1 )( x + 5 ) x + 1 = x + 5 . Therefore, lim x →− 1 x 2 + 6 x + 5 x + 1 = lim x →− 1 x + 5 = 4 .