Therefore, lim
= 2.
→∞
74
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CHAPTER 2 LIMITS
The sequence appears to converge to 0 75 Assume the limit exists so that
28.
⇒
5
0.7654
lim
+1
= lim
2
0.3333
6
0.7449
lim
+1
= lim
3 4
0.8889 0.7037
7 8
0.7517 0.7494
1
2.0000
5
17.0000
2
3.0000
6
33.0000
3 4
5.0000 9.0000
7 8
65.0000 129.0000
1
1.0000
2
2.2361
3
3.3437
4
4.0888
5
4.5215
6
4.7547
7 8
4.8758 4.9375
1
1.0000
2
3.0000
3
1.5000
4
2.4000
5
1.7647
Therefore, if the limit exists it will be either −3 or 2, but since all terms of the sequence are
6
2.1702
positive, we see that lim
7 8
1.8926 2.0742
1
3.0000
2
5.0000
3
3.0000
4
5.0000
5
3.0000
6
5.0000
7 8
3.0000 5.0000
→∞
→∞
1−
→∞
then
1 3
2.0000
→∞
=
= 1−
1
+1
⇒
1 3
= 1−
1 3
⇒
=3 4
= 34 .
Therefore, lim
→∞
29.
The sequence is divergent.
30.
The sequence appears to converge to 5 Assume the limit exists so that √ √ lim = then +1 = 5 ⇒ lim 5 +1 = lim +1 = lim →∞
→∞
√ = 5
⇒
→∞
2
=5
⇒
( − 5) = 0
⇒
→∞
= 0 or
⇒
=5
Therefore, if the limit exists it will be either 0 or 5. Since the first 8 terms of the sequence appear to approach 5, we surmise that lim
= 5.
→∞
31.
32.
The sequence appears to converge to 2 Assume the limit exists so that lim
→∞
=
+1
6 1+
= lim
=
→∞
⇒
2
then
+ −6=0
→∞
+1
= ⇒
6 1+
⇒
lim
→∞
( − 2)( + 3) = 0
= 2.
The sequence cycles between 3 and 5 hence it is divergent.
+1
⇒
= lim
→∞
6 1+
= −3 or
⇒ =2