2.3 Inequalities and Limits 77
Example 2.3.8 lim efFi, = l. (Assume the existence^9 of efri,.)
n-+oo 'f1Yn
Proof. Vn E N, let an = efFi, - 1; i.e., efFi, =an+ l.
By the Algebra of Limits theorem, it suffices to prove that an -t 0. Now,
Vn EN, an ~ 0 (prove). Then, from the equation above, we have
n = (1 + anr, so by Exercise i.3.15,
n > ~n(n - l)a;; that is,
2 > (n - l)a;, or
2
--> a^2 (if n ~ 2).
n-l n
2
Hence, 0 <a;< ---t 0.
- n-l
Therefore, by the second squeeze principle, a; -t 0. Then, by the Algebra
of Limits theorem, .;a;: -t 0. That is, an -t 0. D
Example 2.3.9 For any fixed c > 0, lim y'C = l.
n-+oo [. ~
Proof. Case 1 ( c > 1): By the Archimedean property, 3 no E N 3 no > c.
Then n ~ n 0 =? efFi, > y'C. Thus, \fn EN,
n ~ n 0 =? 1 < y'C < efFi, -t 1 (by Example 2.3.8).
Therefore, by the first squeeze principle, y'C -t l.
Case 2 ( c = 1): Trivial.
Case 3 (0 < c < 1): Then, ~ > 1, so by Case 1, lim n{f -t l; that is,
c n-+ooVc
1
nf;. -t 1.
yC
1
By the Algebra of Limits theorem, this implies that y'C -t l = l. D
- Examples 2 .3.8 and 2.3.9 make use of the nth root function f(x) = ::tX. This function
will be defined rigorously in Section 5.5, specifically, in Exercise 14 of that section. In the
meantime, we shall assume its familiar algebraic properties.