Limits
Roughly speaking, lim an = L means that an approaches L as n grows. Here is the
n→∞
precise definition of limit.
Definition 1 (Limit)
(a) A sequence of real numbers is a rule which assigns to each natural number n a real
number an . We’ll denote it a1 , a2 , a3 , · · · or an n∈IN .
(b) Let L ∈ IR and an n∈IN be a sequence of real numbers. Then lim an = L if
n→∞
∀ ε > 0 ∃ N ∈ IN such that
an − L < ε whenever n ≥ N
The sequence is said to converge to L.
(c) A sequence is said to diverge if it does not converge to any real number.
Remark 2 Here is what that definition of “ lim an = L” says. Suppose you concentrate
n→∞
on a tiny interval centred on L. The interval can be arbitrarily small. (That’s the ∀ ε > 0
in the definition — the interval is (L − ε, L + ε).) Then as along as n is big enough an is
in the interval. (That’s the ∃ N ∈ IN in the definition — if n ≥ N , a is in the interval.)
Example 3 In Example 2 of the notes “A Little Logic” we saw that the statement
∀ ε > 0 ∃ N ∈ IN such that
1
n
< ε whenever n ≥ N
is true. This statement implies that the statement
∀ ε > 0 ∃ N ∈ IN such that
is true for an =
Definition 1, so
1
n
an − L < ε whenever n ≥ N
and L = 0. Of course (1) is exactly the definition of lim an = L in
n→∞
lim 1
n→∞ n
c Joel Feldman.
(1)
2014. All rights reserved.
=0
September 23, 2014
Limits
1
Example 4 In this example, we consider lim (−1)n . The sequence an = (−1)n
n→∞
n
n∈IN
is
−1, 1, −1, 1, −1, 1, −1, · · ·. We would guess that lim (−1) does not exist. So we fix any
n→∞
real number L and show that lim (−1)n cannot be L. To do so, let U be the statement
n→∞
∀ ε > 0 ∃ N ∈ IN such that
(−1)n − L < ε whenever n ≥ N
We wish to show that U is false. To do so, we split it up into bite sized pieces, working
from right to left, just as we did in the notes “A Little Logic”. Precisely, we let (see (2)
below)
◦ S(N, ε) be the statement “(−1)n − L < ε whenever n ≥ N ”, and
◦ T (ε) be the statement “∃ N ∈ IN such that S(N, ε)” or
∃ N ∈ IN such that (−1)n − L < ε whenever n ≥ N
◦ Then U is the statement “∀ ε > 0 T (ε)”.
U
z
z
}|
{
T (ε)
}|
{
S(N,ε)
z
}|
{
n
∀ ε > 0 ∃ N ∈ IN such that (−1) − L < ε whenever n ≥ N
(2)
We analyze U using the same three steps as in Example 2 of “A Little Logic”.
◦ Step 1: We find all N ’s and ε’s for which S(N, ε) is true. Fix any ε > 0 and any
N ∈ IN. The statement S(N, ε) is true if all values of (−1)n , with n ≥ N , lie in the
interval (L − ε, L + ε). Regardless of what N is, the set of all values of (−1)n with
n ≥ N is exactly {−1, 1}. So S(N, ε) is true if and only if both −1 and 1 are contained
in the interval (L − ε, L + ε). That is, S(N, ε) is true if and only if
L − ε < −1 < 1 < L + ε
Now, regardless of what L is, the distance from L − ε to L + ε is 2ε. As the distance
from −1 to 1 is 2. If ε < 1, the interval (L − ε, L + ε) is shorter than the distance
between −1 and 1 and so cannot contain both. So if ε < 1, S(N, ε) is false for all
N ∈ IN.
◦ Step 2: Because S(N, ε)is false for all N ∈ IN when ε < 1, T (ε) is false for all ε < 1.
◦ Step 3: We conclude that U is false since, as we have just seen, T (ε) is false for at
least one ε > 0. For example T 12 is false.
In conclusion, (−1)n has no limit as n → ∞.
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
2
Example 5 In this example, we consider lim n. The sequence an = n n∈IN is 1, 2, 3,
n→∞
4, 5, · · ·. We would guess that lim n does not exist. So we fix any real number L and
n→∞
show that lim n cannot be L. That is, we show that the statement U in
n→∞
U
z
}|
{
∀ ε > 0 ∃ N ∈ IN such that n − L < ε whenever n ≥ N
|
{z
}
(3)
S(N,ε)
{z
|
T (ε)
}
is false.
◦ Step 1: We find all N ’s and ε’s for which S(N, ε) is true. Fix any ε > 0 and any
N ∈ IN. The statement S(N, ε) is true if all values of n, with n ≥ N , lie in the
interval (L − ε, L + ε). But, for any N ∈ IN and any ε > 0 and any L ∈ IR there are
certainly n’s with n ≥ N that obey n > L + ε. So S(N, ε) is false for all N ∈ IN and
ε > 0.
◦ Step 2: Because S(N, ε) is false for all N ∈ IN when ε > 0, T (ε) is false for all ε > 0.
◦ Step 3: We conclude that U is false since, as we have just seen, T (ε) is false for at
least one ε > 0.
In conclusion, n has no limit as n → ∞.
To this point it is logically possible that a sequence has more than one limit. The
following theorem rules that out and also generalises Example 5.
Definition 6 (Bounded) A sequence an
A > 0 such that |an | ≤ A for all n ∈ IN.
n∈IN
is said to be bounded if there exists an
Theorem 7
(a) A sequence has at most one limit.
(b) If a sequence converges, it is bounded.
Proof: (a) We’ll prove this by contradiction. Suppose that the sequence an n∈IN converges both to b and to c and that b 6= c. By definition, given any ε > 0 (we’ll pick a
specific ε shortly),
∃ Nb ∈ IN such that
∃ Nc ∈ IN such that
c Joel Feldman.
2014. All rights reserved.
an − b < ε whenever n ≥ Nb
an − c < ε whenever n ≥ Nc
September 23, 2014
Limits
3
So if n is bigger than both Na and Nb , an is simultaneously within a distance ε of b and
within a distance ε of c. Consequently
|b − c| = |b − an + an − c| ≤ |b − an | + |an − c| < ε + ε = 2ε
That is, the distance from b to c is less than 2ε. This is true for any ε > 0. In particular,
it is true for ε = 31 |b − c|. So
|b − c| < 2ε = 23 |b − c| =⇒
1
3 |b
− c| < 0
which is impossible.
(b) Suppose that the sequence an
n∈IN
converges to L. By definition
∀ ε > 0 ∃ N ∈ IN such that
an − L < ε whenever n ≥ N
In particular, choosing ε = 1, there is an N ∈ IN such that
n ≥ N =⇒ |an − L| < 1 =⇒ |an | = an − L + L ≤ |an − L| + |L| < 1 + |L|
This tells us that all an ’s with index n greater than N are bounded by 1 + |L|. Without
the restriction to n ≥ N , we still have
|an | ≤ max |a1 | , |a2 | , · · · , |a|N−1 , 1 + |L|
Evaluating limits by directly verifying Definition 1 is OK for very simple sequences
but it gets ugly and hard
√ very quickly. Try applying Definition 1 directly to show that,
1+1/n+2/n2
for example, limn→∞ 3+4/n+5/n2 = 31 . It is much more efficient to have a list of simple
limits that we already know together with a toolbox that allows us to build complicated
limits out of simple ones. As examples of known simple limits, it is trivial that, for any
a ∈ IR,
lim a = a
n→∞
and we have already seen, in Example 3, that
lim 1
n→∞ n
=0
The following few theorems provides a toolbox and then we’ll get some more simple examples.
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
4
Theorem 8 (The Arithmetic of Limits) Let a, b, c ∈ IR, Ñ ∈ IN, and an
bn n∈IR , and cn n∈IR be sequences of real numbers. Assume that
lim an = a
n→∞
lim bn = b
n→∞
n∈IR
,
lim cn = c
n→∞
and that c and all of the cn ’s are nonzero. Then
(a)
lim an + bn = a + b
n→∞
(b)
(c)
lim an bn = ab
n→∞
lim 1
n→∞ cn
=
1
c
(d) If an ≤ bn for all n ≥ Ñ , then a ≤ b.
Warning 9 The assumption that an < bn for all n does not imply that lim an < lim bn .
A counterexample is provided by an = 0, bn =
1
,
n
n→∞
n→∞
a = 0, b = 0.
Proof of Theorem 8: We’ll start by writing out our given data. Note that the ε
and N in “∀ ε > 0 ∃ N ∈ IN such that S(N, ε)” are dummy variables, just as x is a
R1
dummy variable in 0 x dx. You may replace ε and N by whatever symbols you like. The
hypotheses of this theorem say
∀ εa > 0 ∃ Na ∈ IN such that
∀ εb > 0 ∃ Nb ∈ IN such that
∀ εc > 0 ∃ Nc ∈ IN such that
an − a < εa whenever n ≥ Na
an − a < εb whenever n ≥ Nb
an − a < εc whenever n ≥ Nc
(4)
(5)
(6)
We also know that any convergent sequence is bounded. So there exist constants A, B > 0
such that
|an | ≤ A and |bn | ≤ B
∀ n ∈ IN
(a) We are to prove that
∀ ε > 0 ∃ N ∈ IN such that |an + bn − a − b| < ε whenever n ≥ N
So pick any ε > 0. We must prove that there is an n ∈ IN such that
|an + bn − a − b| < ε whenever n ≥ N
Observe that
|an + bn − a − b| = (an − a) + (bn − b) ≤ |an − a| + |bn − b|
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
5
We want the right hand side to be smaller than ε. So we apply (4) and (5) with εa = εb =
and conclude that
∃ Na ∈ IN such that |an − a| <
ε
2
ε
2
ε
2
whenever n ≥ Na
∃ Nb ∈ IN such that |bn − b| < whenever n ≥ Nb
Choose N = max Na , Nb . Then whenever n ≥ N we have both n ≥ Na and n ≥ Nb so
that
|an + bn − a − b| ≤ |an − a| + |bn − b| < 2ε + 2ε = ε
(b) We are to prove that
∀ ε > 0 ∃ N ∈ IN such that |an bn − ab| < ε whenever n ≥ N
So pick any ε > 0. We must prove that there is a N ∈ IN such that
|an bn − ab| < ε whenever n ≥ N
Observe that
|an bn − ab| = (an − a + a)bn − ab = (an − a)bn + a(bn − b) ≤ |an − a| B + |a| |bn − b|
ε
We want the right hand side to be smaller than ε. So we apply (4) and (5) with εa = 2B
ε
(the “+1” in the denominator is there just to make sure that we are not
and εb = 2|a|+1
dividing by 0) and conclude that
∃ Na ∈ IN such that |an − a| <
∃ Nb ∈ IN such that |bn − b| <
ε
whenever n ≥ Na
2B
ε
2|a|+1 whenever n ≥ Nb
Choose N = max Na , Nb . Then whenever n ≥ N we have both n ≥ Na and n ≥ Nb so
that
|an bn − ab| ≤ |an − a| B + |a| |bn − b|
<
<
ε
ε
2B B + |a| 2|a|+1
ε
+ 2ε = ε
2
(c) We are to prove that
∀ ε > 0 ∃ N ∈ IN such that c1n − 1c < ε whenever n ≥ N
So pick any ε > 0. We must prove that there is a N ∈ IN such that
c Joel Feldman.
1
cn
− 1c < ε whenever n ≥ N
2014. All rights reserved.
September 23, 2014
Limits
6
Observe that
1
− 1 = |cn − c|
cn
c
|cn c|
To get an upper bound on this, we need a lower bound on the denominator. This denominator, |cn c| varies with n, but, for large n, it is close to c2 . In particular, applying (6)
|c|
5|c|
with εc = |c|
2 , for example (we could just as well use 10 or 6 ), we conclude that
∃ Nc ∈ IN such that |cn − c| <
|c|
2
whenever n ≥ Nc
and then
|c|
|c|
|cn − c|
|cn − c|
=
=⇒
≤2
2
2
|cn c|
c2
n ≥ Nc =⇒ |cn | = |c + (cn − c)| ≥ |c| − |cn − c| > |c| −
To get this smaller than ε, we apply (6) again, this time with εc =
∃ Ñc ∈ IN such that |cn − c| <
c2
ε
2
c2
2 ε,
conclude that
whenever n ≥ Nc
and choose N = max{Nc , Ñc }. Then whenever n ≥ N we have both n ≥ Nc and n ≥ Ñc
so that
c2
1
ε
|c
−
c|
1
n
− ≤2
< 2 22 = ε
cn
2
c
c
c
(d) We’ll prove this by contradiction. Assume that an ≤ bn for all n, but that the limit
b < a. Then (4) and (5) with εa = εb = |a−b|
imply that
2
∃ Na ∈ IN such that |an − a| <
∃ Nb ∈ IN such that |bn − b| <
|a−b|
2
|a−b|
2
whenever n ≥ Na
whenever n ≥ Nb
So if n ≥ max{Na , Nb }, we have
>0
z }| {
a n − bn = a n − a + a − b + b − bn > 0
| {z }
| {z }
>−
|a−b|
2
>−
|a−b|
2
That is an > bn , which is a contradiction when n ≥ Ñ .
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
7
Definition 10 (Monotonic Sequences) A sequence an n∈IN of real numbers is said
to be monotonic if either a1 ≥ a2 ≥ a3 ≥ · · · or a1 ≤ a2 ≤ a3 ≤ · · · .
Theorem 11 (Monotonic Sequences) Any bounded monotonic sequence of real numbers converges.
Proof: Let’s assume that the sequence an n∈IN of real numbers is bounded and obeys
a1 ≤ a2 ≤ a3 ≤ · · · . The other case is similar. Then the set A = an n ∈ IN is
a nonempty, bounded set of real numbers. So a = sup A exists by the least upper bound
axiom. We’ll prove that lim an = a.
n→∞
Let ε > 0. Since a is the least upper bound of A, there must be an element of A
that is strictly bigger than a − ε. Otherwise a − ε is an upper bound for A. Suppose that
element has index N . That is aN > a − ε. Then
n ≥ N =⇒ a − ε < aN ≤ an ≤ a =⇒ |an − a| < ε
Theorem 12 (Squeeze Theorem) Let an
of real numbers that obey:
n∈IN
, bn
n∈IN
and cn
n∈IN
be sequences
(i) There is an n0 ∈ IN such that an ≤ bn ≤ cn for all n ≥ n0 .
(ii) lim an = lim cn = L.
n→∞
n→∞
Then lim bn = L
n→∞
Proof:
On a problem set.
Now for a collection of simple examples. (Well, at least some of them are simple.)
Theorem 13 (Examples)
1
p
n→∞ n
(a) If p > 0, then lim
(b) If p > 0, then lim
n→∞
√
n
=0
p=1
(c) If |x| < 1, then lim xn = 0.
n→∞
α
(d) For any α ∈ IR and b > 1, lim nbn = 0.
n→∞
√
(e)
lim n n = 1
n→∞
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
8
Proof: (a) Let ε > 0. Choose N to be any natural number obeying N >
N p > 1ε and
n ≥ N =⇒ n1p ≤ N1p < ε
1 1/p
.
ε
Then
(b) First consider the case that p ≥ 1. By Problem Set 2, question #2(b),
0≤
√
n
1
n (p
p−1 ≤
− 1)
The right hand side converges to 0. So, by the squeeze theorem, lim
n→∞
√
n
Next consider 0 < p < 1. Then
lim
n→∞
√
n
1
p = lim q
n→∞
n
1
=
1
p
lim
n→∞
q
n
1
p
p − 1 = 0.
=1
(c) Let ε > 0. Observe that
n
√
x − 0 = |x|n < ε ⇐⇒ |x| < n ε
We have already seen, in part (b), that lim
n→∞
√
such that n ε > |x| for all n ≥ N .
√
n
ε = 1. So, as |x| < 1, there is an N ∈ IN
(d) The case α ≤ 0 is trivial, since then nα ≤ 1. So assume that α > 0. Write an =
Then
(1 + 1/n)α
an+1 = an
b
Pick any 1 < b̃ < b and observe that
α
1 + n1 ≤ b̃ ⇐⇒ 1 +
≤ b̃1/α ⇐⇒ n ≥ b̃1/α − 1
−1
So if N is any natural number bigger than b̃1/α − 1
,
1
n
nα
.
bn
−1
′
0 ≤ aN+1 ≤ aN bb
≤ aN
..
.
b′ 2
b
0 ≤ aN+ℓ ≤ aN+ℓ−1 bb ≤ aN
b′ ℓ
b
′
0 ≤ aN+2 ≤ aN+1 bb
..
..
.
.
′
Since 0 <
b′
b
< 1, lim
ℓ→∞
b′ ℓ
b
= 0 and so lim an = 0 by the squeeze theorem.
n→∞
(e) On a problem set.
c Joel Feldman.
2014. All rights reserved.
September 23, 2014
Limits
9