On the other hand x n is a decreasing function if x 1

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On the other hand x n is a decreasing function if x < 1, and must there- fore tend to a limit or to -∞ . Since x n is positive the second alternative may be ignored. Thus lim x n = l , say, and as above l = xl , so that l must be zero. Hence lim x n = 0 (0 < x < 1) . Example. Prove as in the preceding example that (1 /x ) n tends to + if 0 < x < 1, and deduce that x n tends to 0. We have finally to consider the case in which x is negative. If - 1 < x < 0 and x = - y , so that 0 < y < 1, then it follows from what precedes that lim y n = 0 and therefore lim x n = 0. If x = - 1 it is obvious that x n oscillates, taking the values - 1, 1 alternatively. Finally if x < - 1, and x = - y , so that y > 1, then y n tends to + , and therefore x n takes values, both positive and negative, numerically greater than any assigned
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[IV : 72] LIMITS OF FUNCTIONS OF A 162 number. Hence x n oscillates infinitely. To sum up: φ ( n ) = x n + ( x > 1) , lim φ ( n ) = 1 ( x = 1) , lim φ ( n ) = 0 ( - 1 < x < 1) , φ ( n ) oscillates finitely ( x = - 1) , φ ( n ) oscillates infinitely ( x < - 1) . Examples XXVII. * 1. If φ ( n ) is positive and φ ( n +1) > Kφ ( n ), where K > 1, for all values of n , then φ ( n ) + . [For φ ( n ) > Kφ ( n - 1) > K 2 φ ( n - 2) · · · > K n - 1 φ (1) , from which the conclusion follows at once, as K n → ∞ .] 2. The same result is true if the conditions above stated are satisfied only when n = n 0 . 3. If φ ( n ) is positive and φ ( n + 1) < Kφ ( n ), where 0 < K < 1, then lim φ ( n ) = 0. This result also is true if the conditions are satisfied only when n = n 0 . 4. If | φ ( n + 1) | < K | φ ( n ) | when n = n 0 , and 0 < K < 1, then lim φ ( n ) = 0. 5. If φ ( n ) is positive and lim { φ ( n + 1) } / { φ ( n ) } = l > 1, then φ ( n ) + . [For we can determine n 0 so that { φ ( n + 1) } / { φ ( n ) } > K > 1 when n = n 0 : we may, e.g. , take K halfway between 1 and l . Now apply Ex. 1.] 6. If lim { φ ( n + 1) } / { φ ( n ) } = l , where l is numerically less than unity, then lim φ ( n ) = 0. [This follows from Ex. 4 as Ex. 5 follows from Ex. 1.] 7. Determine the behaviour, as n → ∞ , of φ ( n ) = n r x n , where r is any positive integer. [If x = 0 then φ ( n ) = 0 for all values of n , and φ ( n ) 0. In all other cases φ ( n + 1) φ ( n ) = n + 1 n r x x. First suppose x positive. Then φ ( n ) + if x > 1 (Ex. 5) and φ ( n ) 0 if x < 1 (Ex. 6). If x = 1, then φ ( n ) = n r + . Next suppose x negative. Then * These examples are particularly important and several of them will be made use of later in the text. They should therefore be studied very carefully.
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[IV : 73] POSITIVE INTEGRAL VARIABLE 163 | φ ( n ) | = n r | x | n tends to + if | x | = 1 and to 0 if | x | < 1. Hence φ ( n ) oscillates infinitely if x 5 - 1 and φ ( n ) 0 if - 1 < x < 0.] 8. Discuss n - r x n in the same way. [The results are the same, except that φ ( n ) 0 when x = 1 or - 1.] 9. Draw up a table to show how n k x n behaves as n → ∞ , for all real values of x , and all positive and negative integral values of k . [The reader will observe that the value of k is immaterial except in the special cases when x = 1 or - 1. Since lim { ( n + 1) /n } k = 1, whether k be positive or negative, the limit of the ratio φ ( n + 1) ( n ) depends only on x , and the behaviour of φ ( n ) is in general dominated by the factor x n . The factor n k only asserts itself when x is numerically equal to 1.] 10. Prove that if x is positive then n x 1 as n → ∞ . [Suppose, e.g. , x > 1. Then
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