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 167
divisible by 3. Part of this exercise is to prove that the product of
three consecutive numbers is also divisible by 3.)
30. We will prove this statement using its contrapositive. We will assume
that there exists at least one sequence {c„}^i such that c„ >0 for all n
and L< 0. By definition of limit of a sequence, for every ^ > 0 there
exists an N such that |c„ — L\<8 for all n> N. This is equivalent to
stating L — 8<Cn<L-\- 6 for all n> N. Consider the positive number
8o = -L/2. Because L is the limit of the sequence, there exists a
corresponding M^Q such that L — (L/2)<Cn<L + (—L/2), for all
n>MsQ. Therefore,
— <c„<-, for all n>Ms,.
The numbers 3L/2 and L/2 are negative. So we reached the
conclusion that c„<0 for all n>MsQ. This contradicts the statement
that c„ > 0 for all n. Therefore, the limit of a sequence of positive
numbers cannot be negative. We cannot say that the limit of a
sequence of positive numbers has to be positive, because it could be 0.
Indeed, consider the sequence {l/n}^j. Every term of the sequence is
positive, but the limit of this sequence is 0.
31. To prove that limx->3/(x) = V^, one needs to prove that for every
given 6:>0 there exists a 5>0 such that \f{x) — V^| <£ for all x with
|x-3|<5. Observe that
/(x)-V3U|V3^-V3U
(v^-V3)(V3^ + ^/3)
(v^+^/3)
Therefore, \f{x) — A/3| = |x - 3| r^^ R- HOW large is the factor J^ j-
for X relatively close to 3? If |x — 3| < 1 (this is an arbitrary choice), then
2<x<4and V2<V^<A/4 = 2. Thus, V2 +V3<V^N/3<2 +V3,
or2V2<v^ +V3<4.Therefore,(l/(2x/2))> (1/(V3^ +V3))> (1/4),
and \f{x) - \/3| = |x - 3| -j^^
^ 27! '^ ~ ^1* ^^^ ^^'^^ of this expression
will be smaller than the given £>0if^|x-3|<£; that is, if
|x - 3| <2 V2£. Thus, choose 8 = min{ 1,2^2^}.
32. Existence: Because ad — bc^ 0, at least two of these four numbers
are not equal to 0. Without loss of generality, we will assume
a^O. From the first equation we obtain x = (e — by)/a. We can
substitute this formula for x into the second equation to obtain