2. Use the method of Example 3.4.3(b) to show that if 0

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b. lim (c/n)
= 1 for c> 1.
This limit has been obtained in Example 3.1.11(c) for c > 0, using a
rather ingenious argument. We give here an alternative approach
for the case c> 1. Note that if % = c¹, then zn > 1 and
Zn+1 <zn for all n = N. (Why?) Thus by the Monotone
Convergence Theorem, the limit x = lim (zn) exists. By Theorem
3.4.2, it follows that x = lim (2n). In addition, it follows from the
relation
Z2n = c1/2n = (c²/n) 1/2
and Theorem 3.2.10 that
1/2
z= lim (2n) = (lim (2n)) 1/2 = 1/2
Therefore we have z² =z whence it follows that either
x = 0 or x = 1. Since zn 1 for all nЄ N, we deduce that x = 1.
We leave it as an exercise to the reader to consider the case
0<c<1.
Transcribed Image Text:b. lim (c/n) = 1 for c> 1. This limit has been obtained in Example 3.1.11(c) for c > 0, using a rather ingenious argument. We give here an alternative approach for the case c> 1. Note that if % = c¹, then zn > 1 and Zn+1 <zn for all n = N. (Why?) Thus by the Monotone Convergence Theorem, the limit x = lim (zn) exists. By Theorem 3.4.2, it follows that x = lim (2n). In addition, it follows from the relation Z2n = c1/2n = (c²/n) 1/2 and Theorem 3.2.10 that 1/2 z= lim (2n) = (lim (2n)) 1/2 = 1/2 Therefore we have z² =z whence it follows that either x = 0 or x = 1. Since zn 1 for all nЄ N, we deduce that x = 1. We leave it as an exercise to the reader to consider the case 0<c<1.
2. Use the method of Example 3.4.3(b) to show that if 0 <c<1, then
lim (c¹/n) = 1.
Transcribed Image Text:2. Use the method of Example 3.4.3(b) to show that if 0 <c<1, then lim (c¹/n) = 1.
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