# X n is bounded from the squeeze theorem for sequences

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x n ) is bounded. From the squeeze theorem for sequences (Theorem 2.9, part (ii)), it follows that f ( x n ) g ( x n ) 0. Hence from Theorem 3.6, we get lim x a f ( x ) g ( x ) = 0. 2 8, page 70. ( ) First assume that lim x →∞ f ( x ) = L and let x n ( a, ) such that x n → ∞ as n → ∞ . We want to show that f ( x n ) L . Let ± > 0. Since lim x →∞ f ( x ) = L , there exists M R such that for any x > M , | f ( x ) - L | < ± . Because x n → ∞ , there exists a rank N 0 N such that n N 0 , x n > M . Combining these, it follows that n N 0 , | f ( x n ) - L | < ± . Thus f ( x n ) L . ( ) Conversely, now assume that f ( x n ) L for any sequence x n ( a, ) such that x n → ∞ , and we’d like to prove that lim x →∞ f ( x ) = L . We do this using a proof by contradiction. Suppose that f ( x ) does not converge to L as x approaches a . Then there exists an ± 0 > 0 such that for any M R , there exists a real number x M (which depends on M ) such that x M > M and | f ( x M ) - L | ≥ ± 0 . In particular, for each natural number n > a (taken as M in the above), we’ll ﬁnd a real number x n , such that x n > n and | f ( x n ) - L | ≥ ± 0 . The inequality x n > n for any n > a implies that the sequence x n → ∞ , thus by assumption f ( x n ) L as n → ∞ . Thus | f ( x n ) - L | < ± 0 , for n suﬃciently large. But this is in contradiction with | f ( x n ) - L | ≥ ± 0 , for any n N , n > a . 2
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