Download MATH1023 Calculus I, 2016

MATH1023 Calculus I, 2016-17 Fall
1.1.55
1 The sequence
√
n! + 1 −
Hw 2
√
√
√
n! − 1 is a subsequence of n + 1 − n − 1. And
√
√
(n + 1) − (n − 1)
√
n + 1 − n − 1 = lim √
=0
n→∞
n→∞
n+1+ n−1
√
√
So the subsequence has the same limit. That is limn→∞ n! + 1 − n! − 1 = 0.
lim
1
1
1
1
2 The sequence (n!) n! is a subsequence of n n . And limn→∞ n n = 1. So limN →∞ (n!) n! = 1.
1
3 The sequence ((n + 1)!) n! satisfies
1
1
1
1
(n!) n! ≤ ((n + 1)!) n! ≤ (n!) n! · (n + n) n
1
1
Because limn→∞ (n!) n! = 1 and limn→∞ (2n) n = 1, by the Arithmetic rule and the Sandwich rule,
1
limn→∞ ((n + 1)!) n! = 1.
1
4 The sequence ((n + (−1)n )!) n satisfies
1
1
1 ≤ ((n + (−1)n )!) n ≤ ((n + 1)!) n!
1
By limn→∞ ((n + 1)!) n! = 1 and the Sandwich rule, we have that
1
lim ((n + (−1)n )!) n = 1
n→∞
1
5 The sequence (n!) (n+1)! satisfies
1
1
1 ≤ (n!) (n+1)! ≤ (n!) n!
1
By limn→∞ (n!) n! = 1, and the Sandwich rule, we have that
1
lim (n!) (n+1)! = 1
n→∞
2
6 The sequence (2n
−1
2
1
1
+ 3n ) n2 is a subsequence of (2n−1 + 3n ) n . And
1
3 ≤ (2n−1 + 3n ) n = 3 ·
( ( )n
) n1
1
1 2
+1
≤ 3 · 2n
2 3
1
Then by limn→∞ 3 · 2 n and the Sandwich rule, we have that
lim (2n
n→∞
1.1.56
2
−1
2
1
+ 3 n ) n2 = 3
1 The even subsequence is a2n = 2 and the odd sequence is a2n+1 =
converge to different limit so the original sequence does not converge.
1
2.
The two subsequences
2 The even subsequence is a2n = (2n) 2n and the odd subsequence is a2n+1 = (2n + 1)− 2n+1 =
1
. Both subsequences converge to 1 and the union of {a2n } and {a2n+1 } is the original
1
1
1
(2n+1) 2n+1
sequence {an }. So the original sequence {an } converges to 1, too.
1
3 The even subsequence is a2n = 2n and the odd sequence is a2n+1 = 2n+1
. So limn→∞ a2n goes to
infinity while limn→∞ a2n+1 = 0. The two limits do not equal to each other so the original sequence
does not converge.
2n+3
. Its limit is 1. The odd subsequence is a2n+1 = −2n+2
4 The even subsequence is a2n = 2n−2
2n+3 . Its
limit is −1. The two subsequences have different limits. So the original sequence does not converges.
1
2
2
(2n)
(2n+1)
1
5 The even subsequence is a2n = (2n)
3 −1 . Its limit is 2 . The odd subsequence is a2n+1 = − (2n+1)3 −1 .
Its limit is − 21 . The two subsequences have different limits. So the original sequence does not
converge.
√ √
√
6 The even subsequence is a2n = 2n( 2n + 1 − n). Its limit is 12 . The odd subsequence is
√
√
√
a2n+1 = 2n + 1( 2n − 2n + 1). Its limit is − 12 . The two subsequences have different limits. So
the original sequence does not converge.
1
7 The even subsequence is a2n = (22n + 32n ) 2n . Its limit is 3. The odd subsequence is a2n+1 =
1
(22n+1 + 32n+1 ) 2n+1 . Its limit is 3. The two subsequences have the same limit. And the union of
the odd subsequence and the even subsequence is the original sequence. So the limit of the original
sequence exists and it equals to 3.
1
8 The even subsequence is a2n = (22n + 32n ) 2n . Its limit is 3. The odd subsequence is a2n+1 =
1
1
) 2n+1 . Its limit is 2. The two subsequences have different limits. So the original
(22n+1 + 32n+1
sequence does not converge.
√
9 Consider 3 subsequences: a3n = tan 3nπ
= tan nπ = 0, a3n+1 = tan (3n+1)π
= tan π3 = 3.
3
3
√
a3n+2 = tan (3n+2)π
= tan 2π
3
3 = − 3. The subsequences have different limits implies that the
original sequence does not converge.
√
(6n+1)π
10 Consider two subsequences: a6n = sin 6nπ
= − sin π3 = − 23 . These
3 = 0, and a6n+1 = − sin
3
two subsequences converge to different limits. So the original sequence {an } does not converge.
12nπ
11 Consider two subsequences: a12n = sin 12nπ
= 0 and a12n+1 = sin (12n+1)π
cos (12n+1)π
=
2 cos 3
2
3
π
π
1
sin 2 cos 3 = 2 . These two subsequences converge to different limits. So the original sequence {an }
does not converge.
12 Consider two subsequences: a12n =
√
(12n+1)·
2
12n sin 12nπ
3
12n cos 12nπ
2 +2
= 0 and a12n+1 =
(12n+1)π
3
(12n+1)π
+2
2
(12n+1) sin
(12n+1) cos
=
3
2
. The sequence {a12n+1 } diverges. These two subsequences don’t converge to the same
limits. So the original sequence {an } does not converge.
13 Using arithmetic rule, and the sandwich rule we can get the limit of this sequence is 1.
1.1.57 Assume 0 < a < π, For any positive integer k, the interval [kπ − π2 , kπ + π2 ] of length π contains the
following interval of length a
[
a
a]
[ak , bk ] = kπ − , kπ +
2
2
For even k, we have cos x ≥ cos a2 > 0 on [ak , bk ]. For odd k, we have cos x ≤ − cos a2 < 0 on [ak , bk ].
Since the arithmetic sequence a, 2a, 3a, · · · has increment a, which is the length of [ak , bk ], we must
have nk a ∈ [ak , bk ] for some natural number nk . Then cos(n2k a) is a subsequence of cos na satisfying
cos n2k a ≥ cos a2 > 0 and cos n2k+1 a is a subsequence satisfying cos n2k ≤ − cos a2 < 0. Therefore the
two subsequences cannot converge to the same limit. As a result, the sequence cos na diverges.
For general a, we have a = 2N π ± b for an integer N and b satisfying 0 < b < π. Then we have
cos na = ± cos nb. We have shown that cos nb diverges, so that cos na diverges.
So cos na converges if and only if a = 0.
1.1.58 If yn is bounded, then we may assume that m < yn < M . The sequence xn is the union of two
subsequences x′k and x′′k satisfying all x′k ≥ 1 and all x′′k ≤ 1. By Proposition 1.1.6, the assumption
limn→∞ xn = 1 implies that limk→∞ x′k = limk→∞ x′′k = 1.
2
x′k ≥ 1 implies x′ k ≤ x′ kn ≤ x′ k . And x′′k ≤ 1 implies x′′ k ≥ x′′ kn ≥ x′′ k . By Example 1.1.21
m
y
M
m
y
M
lim x′ k = 1
m
k→∞
lim x′ k = 1
M
k→∞
lim x′′ k = 1
m
k→∞
lim x′′ k = 1
M
k→∞
and the Sandwich rule, limn→∞ x′ kn = limn→∞ x′′ kn = 1. Since the sequence xynn is the union of two
y
y
subsequences x′ kn and x′′ kn , by Proposition 1.1.6 again, we get limn→∞ xynn = 1.
( )p
1.1.59 If limn→∞ xn = l > 0, then by Arithmetic rule, limn→∞ xln = 1. Then by Example 1.1.21, limn→∞ xln =
1. Use Arithmetic rule again, we have that limn→∞ xpn = lp .
y
1.2.1
y
2
2
n − 1
n − 1 − n2 − 1 2
2
=
n2 + 1 ≤ n
n2 + 1 − 1 = n2 + 1
for sufficiently big n.
Then for any ϵ > 0, take N = 2ϵ . Then for any n > N , we have
2
n − 1
2
n2 + 1 − 1 ≤ n < ϵ
Then by the definition of limit, limn→∞
n2 −1
n2 +1
= 1.
1.2.2 The range of the height of the wall is [2.7, 3.3] and the range of the width of the wall is [5.4, 6.6]. So the
area of the wall is in the range of [14.58, 21.78]. So the cost is in the range of [196.83, 294.03].
1.2.5 5x − 3y + 4z = 5(x + 2) − 3(y − 3) + 4(z − 5) + 1. If we want 5x − 3y + 4z to be within ±ϵ of 1, we may
set |x + 2| ≤ 3ϵ , |y − 3| ≤ 3ϵ , |z − 5| ≤ 3ϵ , respectively.
1.2.6 xy − 4 = xy − 2y + 2y − 4 = (x − 2)(y − 2). So we could take the tolerance of x, y near 2 and 2 to be
within ϵ and 1 respectively. Then xy could be within ±ϵ of 4.
1.2.7 x2 − 4 = (x + 2)(x − 2). We could take the tolerance of x near 2 to be min{ 21 , 6ϵ }, then x2 is within ±ϵ
of 4.
1
1
1.2.8 x − 0.5 = |x−2|
|2x| . So we can make sure x is within ±0.1 of 0.5, by taking the tolerance to be 0.2. Then
the percentage tolerance is 0.2
2 = 0.1 = 10%.
3