Solutions for assignment 2
1. Define a sequence {xn } inductively by x1 = 2 and
6
xn+1 =
for n ≥ 1.
7 − xn
(a) Prove that 1 ≤ xn ≤ 2 for all n ≥ 1.
(b) Prove that xn+1 ≤ xn for all n ≥ 1.
(c) Prove that {xn } converges and find the limit, justifying your answer using
results from the lectures.
Solution. (a) We prove this by induction. For the base case n = 1, x1 = 2
satisfies 1 ≤ x1 ≤ 2. Suppose that n ∈ N and 1 ≤ xn ≤ 2. Then 7 − xn satisfies
6
satisfies
5 = 7 − 2 ≤ 7 − xn ≤ 6 = 7 − 1, and hence xn+1 = 7−x
n
6
6
6
≤ ≤ 2.
≤ xn+1 =
6
7 − xn
5
Thus by induction, 1 ≤ xn ≤ 2 for all n ≥ 1.
(b) For n ≥ 1 we have
6
6 − xn (7 − xn )
x2 − 7xn + 6
− xn =
= n
.
7 − xn
7 − xn
7 − xn
For all real x, and in particular x = xn , we have x2 − 7x + 6 = (x − 6)(x − 1).
Since 1 ≤ xn ≤ 2, we have xn − 6 < 0, xn − 1 ≥ 0, and 7 − xn > 0, so
(xn − 6)(xn − 1) ≤ 0, and
xn+1 − xn =
x2n − 7xn + 6
(xn − 6)(xn − 1)
=
≤ 0.
7 − xn
7 − xn
≤ xn for all n ≥ 1.
xn+1 − xn =
But this says xn+1
(c) Part (b) says that the sequence {xn } is nonincreasing, and part (a) implies
that it is bounded below by 1. So the monotone convergence theorem implies
that {xn } converges. Say xn → x as n → ∞.
Since 1 ≤ xn ≤ 2 for all n ≥ 1, part (e) of the algebra of limits implies that
1 ≤ x ≤ 2. Since xn → x, we also have xn+1 → x as n → ∞, and the algebra
of limits implies that
6
6
xn+1 =
→
.
7 − xn
7−x
6
Thus by the uniqueness of limits, x = 7−x
. Solving this equation shows that
x = 1 or x = 6. Since we know that 1 ≤ x ≤ 2, we must have x = 1.
2. Prove that the following sequences converge, find the limits, and justify your
answers using results from the course.
n n3 + n + 6 o
.
(a)
2n3 − 1
1
2
(b)
n 2n2 + (−1)n o
.
n4 + π sin(en )
Solution. (a) We have
n3 + n + 6
=
2n3 − 1
n3 +n+6
n3
2n3 −1
n3
1 + n12 +
=
2 − n13
6
n2
.
We know from our list of standard examples that constant sequences converge,
that n12 → 0, and that n13 → 0. Thus the algebra of limits implies that n62 → 0
and that
1 + n12 + n63
n3 + n + 6
1
=
→ .
1
3
2n − 1
2
2 − n3
(b) We have
2n2 + (−1)n
=
n4 + π sin(en )
2n2 +(−1)n
n4
n4 +π sin(en )
n4
=
2
n2
(−1)n
n4
π sin(en )
n4
+
1+
We consider each of the pieces in this fraction:
• For all n ∈ N, we have −1 ≤ (−1)n ≤ 1. Thus
(−1)n
1
−1
≤
≤ 4.
4
4
n
n
n
1
Since n4 → 0, the algebra of limits implies that −1
→ 0 as well. Thus the
n4
(−1)n
squeeze principle implies that n4 → 0.
• For all real numbers x, we have −π ≤ π sin x ≤ π. Thus for all n ∈ N we
have
π sin(en )
π
−π
≤
≤ 4.
4
4
n
n
n
Since n14 → 0, the algebra of limits (part (b) ) implies that −π
and nπ4 → 0
n4
n
)
→ 0.
as well. Thus the squeeze principle implies that π sin(e
n4
• We have that constant sequences converge, n12 → 0 and n22 → 0 (by the
algebra of limits).
Putting this all together, the algebra of limits gives
n
2
+ (−1)
2n2 + (−1)n
0+0
n2
n4
=
= 0.
n) →
π
sin(e
4
n
n + π sin(e )
1+0
1 + n4
3. Prove from first principles that
x2 − 4
→ −5 as x → 3.
x−4
Solution. Fix > 0. We need to find δ > 0 such that
x2 − 4
0 < |x − 3| < δ =⇒ − (−5) < .
x−4
3
We have
x2 − 4
x2 − 4 + 5(x − 4) x2 + 5x − 24 |x + 8| |x − 3|
− (−5) = .
=
=
x−4
x−4
x−4
|x − 4|
We make the preliminary restriction that |x − 3| < 21 (chosen to keep x away
from 4). Then 2 21 < x < 3 21 , so 10 12 < x + 8 < 11 12 and −1 12 < x − 4 < − 12 .
Thus for |x − 3| < 12 we have |x − 4| > 21 , and
x2 − 4
|x + 8| |x − 3|
− (−5) < ⇐=
<
x−4
|x − 4|
11 12 |x − 3|
⇐=
<
1
2
⇐= 23|x − 3| < .
So δ = min( 23 , 12 ) has the required property.