MATH 101 V2A
February 27th – Practice problems
Hints and Solutions
1. Consider the sequence {an } = sin
nπ
3
.
(a) Show that {an } diverges.
√
√
Solution: Since the sequence only has three distinct terms, 23 , 0 and − 23 , we expect the sequence
to diverge in the sense that lim an DNE and isn’t ±∞. The easiest way to prove this is by using
n→∞
the method of contradiction: suppose instead that {an } converges to some limit a. Then there is
a positive integer N such that for all n ≥ N
a−
Choosing n = 3N , sin
nπ
3
= sin
3N π
3
1
1
< an < a + .
4
4
= sin(N π) = 0, so the above inequality gives us that
a−
1
1
<0<a+ ,
4
4
which tells us that − 14 < a < 41 . Choosing n = 6N + 1, sin nπ
= sin
3
√
3
π
sin 3 = 2 , so the first inequality gives us that
√
3
1
1
a− <
<a+ .
4
2
4
6N π+π
3
= sin 2N π +
Using the fact that − 41 < a < 14 , we get that
√
1 1
3
1
1
1 1
− − <a− <
<a+ < + .
4 4
4
2
4
4 4
√
However,
3
2
> 12 , so we have a contradiction. Therefore the sequence must diverge.
(b) Find two distinct convergent subsequences of {an }, and one divergent subsequence of {an }.
Solution: In the solution above, we noted that sin (3n)π
= 0 and sin (6n+1)π
=
3
3
(3n)π
sin
= {0, 0, 0, . . .},
3
)
( √ √ √
(6n + 1)π
3
3
3
sin
=
,
,
,...
3
2
2
2
are clearly two convergent subsequences. The subsequence
)
√ √
( √ √
3
3
3
3
(3n + 1)π
sin
= −
,
,−
,
,...
3
2
2
2
2
is a divergent subsequence.
√
3
2 .
So,
π
3
=
2. Consider the recursively defined sequence
a1 = 1,
an+1 =
√
6 + an
for n ≥ 1.
(a) Show that the sequence is bounded and monotone increasing by following the steps outlined below.
(i) Show that an+1 ≥ an provided an ≤ 3.
√
Solution: If an = an+1 , then an = 6 + an . So a2n = 6 + an , which means that
√ either an = 3
or an = −2. Therefore, if an ≤ 3, we must have that a2n ≤ 6 + an , or an ≤ 6 + an = an+1 .
(ii) Use mathematical induction to show that an ≤ 3 for all n ≥ 1.
Solution: We are given that a1 = 1 ≤ 3, so it is true in the case n = 1. Now suppose it is true
when n = k for some k, i.e. ak ≤ 3. We need to show that this
√ it is also true
√ implies that
when n = k + 1, i.e. ak+1 ≤ 3. Now, if ak ≤ 3, then ak+1 = 6 + ak ≤ 6 + 3 = 3. This
means that if it true when n = k, then it must also be true when n = k + 1. Since we know
that it is true when n = 1, we therefore get that it is true when n = 2, 3, 4, . . .. Therefore
an ≤ 3 for all n ≥ 1.
(iii) Use (i) and (ii) to conclude that {an } is monotone increasing and bounded by 3.
Solution: From (ii) we know that an ≤ 3 for all n ≥ 1, and so {an } is bounded above by 3
(and clearly below by 0). This, together with (i), also implies that an+1 ≥ an for all n ≥ 1,
so we see that the sequence is monotone increasing.
(b) Use the Monotone Convergence Theorem to conclude that lim an = 3
n→∞
Hint: We did this part in class. (Since we’ve already showed that the sequence is monotone and
bounded, the MCT tells us that the sequence converges to some limit a. Using this, take the limit
of both sides of the defining equation for an+1 and solve for a.)
(c) Calculate the limit of the Fibonacci sequence
a1 = 1,
a2 = 1,
an+2 = an + an+1 for n ≥ 1.
(Hint: lim an 6= 0.)
n→∞
Hint: The Fibonacci sequence diverges. Note that the Fibonacci sequence is unbounded, and therefore
does not satisfy the hypotheses of the MCT. This is important because if we try to find the limit of
2
the Fibonacci sequence by using the same method as in (b) (which uses the MCT!), we would get that
a = 2a and conclude that a = 0, which is very clearly wrong!
3. (a) Show that if lim an = L, then {an } is bounded.
n→∞
Solution: Suppose lim an = L. Then there is a positive integer N such that for every n ≥ N ,
n→∞
L − 1 ≤ an ≤ L + 1.
Hence every term in the sequence {an } is bounded above by
max{a1 , a2 , a3 , . . . , aN −1 , L + 1},
(the biggest number in the (finite) list of numbers {a1 , a2 , a3 , . . . , aN −1 , L + 1}) and bounded below
by
min{a1 , a2 , a3 , . . . , aN −1 , L − 1}.
(the smallest number in the (finite) list of numbers {a1 , a2 , a3 , . . . , aN −1 , L − 1}.) Therefore {an }
is a bounded sequence.
(b) Is it true that if lim f (x) = L, then f is bounded?
x→∞
Solution: No, this is not true. If f (x) =
1
x,
then lim f (x) = 0, but f is unbounded at x = 0 (in
x→∞
fact lim+ f (x) = ∞).
x→0
4. The Sequential Criterion for limits of functions states that the following statements are equivalent:
(i) lim f (x) = L.
x→a
(ii) Given any sequence {an } which converges to a (i.e. lim an = a) and is such that an 6= a for all
n→∞
n, the sequence {f (an )} converges to L (i.e lim f (an ) = L).
n→∞
This theorem provides a concrete connection between the limit of a function and the limits of sequences.
Think about it!!
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