epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Solutions for the Exercises at the End of the Sections and the Review Exercises 149
Thus, 1.25<x2 + l< 13.25, and 6.25< 5(x^ + 1)<66.25; that is,
(1/66.25) <(l/5(x2 + l))< (1/6.25). Combining these estimates with
(^+2)1
the ones for |x + 2| yields ^. z^n <
5(x2+l) l-^
one needs
5.5 _ 22
6.25 = "~ 25- ^. Thus, 1 1 _
(x2+l) 5 -
2| < |§ |x - 2|. To have this expression be smaller than e,
2\<^s. Therefore, let 8 = minimum{1.5, ^s}.
3. Let e>0 be given. We need to prove the existence of a number 8>0
such that if \x-l\<8,
then Kzi
<s. Note that the function
(x^ - l)/(x^ — 1) is undefined at 1 and -1. It can be rewritten as
(x^ - l)/(x2 - 1) = (x - l)(x2 + X + l)/(x - l)(x + 1) = (x2 + X + 1)/
(x + 1) when x ^ 1 and x / -1. Thus, additional algebraic steps yield
x^+x+l _ 3 2x2+2x+2-3x-3
2x2--x-1
=
Because
x+1 2| ~ I 2(x+l)
2(x+l)
2x^ - X - 1 = (2x + l)(x - 1), we ^e
3
X2-1 2 = 2|x-fl| 1-^ ^1
Therefore, an estimate for ^ ^ is needed, knowing that x is in an
interval centered at 1. Because x has to be in an interval centered at 1,
we can choose 0<x<2 {i.e., an interval of radius 1). (This is a
completely arbitrary choice. Other choices will work as well; they will
just yield different results for 8>0. For example, check what would
happen when one uses — l<x<3.) Therefore, l<x+l<3; that is,
l<|xH-l|<3. Then |<K77TT<1, and z < OUTTTT < i Let us now
5<^<1' and i<__^^ ^
consider the factor |2x+l|. Because l<2x-l-l<5, l<|2x + l|<5
Thus, §±|| < f, and <f|x- 1|. For this quantity to be
smaller than the given £>0, one needs |x—l|<i£. Therefore, let
5 = minimum{l,§^}.
4. Let 5 = 0.9. Then |x~2|<5 = 0.9. This implies -0.9<x ~ 2<0.9;
that is, l.l<x<2.9. Therefore, 3.3<3x<8.7. Subtracting 6 yields
-2.7<(3x - 5) - 1 <2.7. So, |(3x - 5) - l| <2.7<4.5.
5. Assume that |x-2|<5<1.5. Then -1.5 < ~-5<x ~ 2<5 < 1.5.
Adding 2 to all the parts of the inequality yields 0.5<2-5<x<2+
8 < 3.5. In turn, multiplying these inequaUties by 3 implies
that 1.5 < 3(2 - 5) < 3x < 3(2 -f 5) < 10.5. Subtracting 6 yields
1.5 - 6<3x - 6< 10.5 - 6 {i.e., -4.5<3x - 6<4.5). Therefore,
|(3x-5)-1|<4.5 whenever |x-2|<5< 1.5.
6. To prove that lim„^oo ^ix = 0 means proving that for every given
> 3n+l
1
s>0 there exists an N such that 0 <8 for all n>N. Because
3n+l
1
3^-0 3n+l (l/(3n + 1)), we need to find values of n that
satisfy the inequality (l/(3n+1))<£. This yields 3n+l>(l/£), or
n>(l/3)({l/6) - 1). Thus, let N = (1 - s)/{3s).