Page 164 Problem 6cde
Using only Definition 16.2, prove the following.
3n+1
n→∞ n+2
(c). lim
= 3.
Solution
5
5
and set
Let ε > 0 be given,
N = ε. Suppose n > N . Then n > ε , so
3n+1
3n+1−(3n+6) −5 5
n+2 − 3 = = n+2 < n < ε. Voilà
n+2
(d) lim
n→∞
sin n
n
n+2
2
n→∞ n −3
< ε, and
1
n
< ε, and
= 0.
Solution
ε > 0 be given, and set N = 1ε . Suppose n > N . Then n > 1ε , so
Let
sin n − 0 ≤ 1 < ε. El fin.
n
n
(e) lim
5
n
= 0.
Solution
Let ε > 0 be given, and set N = max 2, 4ε . Suppose
n > N . Then n > 2, so 0 < n + 2 < 2n, and
4
4
4
n2
n+2
2n
2
0 < 2 < n − 3. Also, n > ε , so n < ε. Therefore, nn+2
2 −3 − 0 = n2 −3 < n2 = n < ε. 2