# Math 112 Solutions For Section 2.3, p. 132

1.
(b) 1. (d) DNE. (f) . (h) DNE. (j) . (k) .
2.
(b) . (d) DNE. (e) 0. (f) . (g) . (h) . (k) 1. (l) .

E.
If we substitute in in the top and the bottom, we get the case. So we look for a factor of in the top and in the bottom. Since and (use polynomial or synthetic division to divide by ), we can cancel the from the top and from the bottom. Thus, .

3.
(a) . (d) . (e) . (f) 2. (g) . (h) (l) DNE. . (k) 0.
4.
(a) 1. (b) 0. (d) 0. (e) 0. (h) DNE. (i) 0. (j) DNE (undefined for ).
5.
Theorem. .

Proof ( ) If as , then . Hence as .

( ) If as , then . Therefore as .

6.
For example, has no limit as , but .

8.
Theorem. If .

Proof. Since as , we can choose close enough to that . Then, given any , we can choose close enought to that . Then, with close enough to , we have .

14.
(a) 0 (b) 0 (c) 0

(d) Undetermined. For example, if and , then , which depends on .

(e) Undetermined. For example, if , then , which depends on .

16.
, and

. Hence

.

18.
(a) bounded, . (b) not bounded. (c) bounded, . (d) not bounded. (e) bounded, . (f) bounded, .

19.
is not bounded, but is bounded for not 0. If exists and is some nonzero number, then is not bounded near 0. Hence if is bounded for all , then either as or does not exist.

20.
(a) . (b) . (c) . (d) The area of an inscribed polygon with n sides is and the area of the circle is .