ALGEBRA OF LIMITS
Proposition. (Algebra of Limits, sequences) Suppose that (an )n and (bn )n are
sequences of reals such that an → a and bn → b. Then:
(1) an + bn → a + b;
(2) an bn → ab;
(3) if all bn are non-zero and b 6= 0, then an /bn → a/b;
Proof.
(1) Let > 0. As an → a, there is N1 ∈ N such that |an − a| < /2 for
all n ≥ N1 . As bn → b, there is N2 ∈ N such that |bn − b| < /2 for all
n ≥ N2 . Let N = max(N1 , N2 ). Then, for all n ≥ N , we have:
|an + bn − (a + b)| = |(an − a) + (bn − b|
≤ |(an − a)| + |(bn − b)|
(triangle inequality)
< /2 + /2
(as n ≥ N1 and n ≥ N2 )
=
as required.
(2) Let > 0. First note that, since (an )n and (bn )n are convergent, they are
bounded; so pick X with X ≥ |an | and X ≥ |bn | for all n, and X ≥ |a|, |b|.
Note that, for all n,
|an bn − ab| = |(an − a)bn + (bn − b)a|
(multiply out)
≤ |an − a||bn | + |bn − b||a|.
As an → a, there is N1 ∈ N so that |an − a| < 2X
for all n ≥ N1 ; similarly
there is N2 ∈ N so that |bn −b| < 2X for all n ≥ N2 . Let N = max(N1 , N2 ).
Then for all n ≥ N we have
|an bn − ab| ≤ |an − a||bn | + |bn − b||a|
<
(|bn | + |a|)
2X
≤
(X + X)
2X
=
(above)
(as n ≥ N )
(choice of X)
as required.
(3) We show that, if b 6= 0 and bn 6= 0 for all n, then 1/bn → 1/b. To get the
full statement, then use part 2.
Let > 0. Since b 6= 0 and bn → b, there is M ∈ N such that |bn − b| <
b/2 for n ≥ M ; so for n ≥ M , |bn | ≥ |b|/2 (by the triangle inequality).
There also exists N1 ∈ N such that |bn − b| < |b|2 /2 for all n ≥ N1 . Let
1
2
ALGEBRA OF LIMITS
N = max(N1 , M ). Then, for n ≥ N , we have:
1
− 1 = b − bn bn
b bbn |b − bn |
=
|b||bn |
|b|2 /2
<
|b||bn |
|b|2 /2
≤
|b|2 /2
= .
(as n ≥ N1 )
(as n ≥ M )
Proposition 0.1. (Algebra of Limits, functions) Suppose that A ⊂ R, a is a limit
point of A, and that f, g : A → R are functions such that limx→a f (x) = l and
limx→a g(x) = m. Then:
(1) limx→a f (x) + g(x) = l + m;
(2) limx→a f (x)g(x) = lm;
(x)
(3) limx→a fg(x)
= l/m if m 6= 0 and g(x) 6= 0 for all x ∈ A.
Proof. This is basically identical to the proofs above, but using δ instead of N . In
fact, I wrote this proof using copy and paste.
(1) Let > 0. As f (x) → l, there is δ1 > 0 such that |f (x) − l| < /2 for
all x ∈ A with 0 < |x − a| < δ1 . As g(x) → m, there is δ2 > 0 such that
|g(x)−m| < /2 for all x ∈ A with 0 < |x−a| < δ2 . Let δ = min(δ1 , δ2 ) > 0.
Then, for all x ∈ A with 0 < |x − a| < δ, we have:
|f (x) + g(x) − (l + m)| = |(f (x) − l) + (g(x) − m)|
≤ |f (x) − l| + |g(x) − m|
(triangle inequality)
< /2 + /2
(choice of δ)
=
as required.
(2) Let > 0. First note that, as f (x) → l and g(x) → m as x → a, they
are bounded near a; there is X ∈ R and δ 0 > 0 such that |f (x)| < X and
|g(x)| < X for |x − a| < δ 0 . Note that, for all x,
|f (x)g(x) − lm| = |(f (x) − l)g(x) + (g(x) − m)l|
(multiply out)
≤ |f (x) − l||g(x)| + |g(x) − m||l|.
As f (x) → l, there is δ1 > 0 so that |f (x) − l| < 2X
for all x ∈ A with
0 < |x − a| < δ1 ; similarly there is δ2 > 0 so that |g(x) − m| < 2X
for all
0
x ∈ A with 0 < |x − a| < δ2 . Let δ = min(δ1 , δ2 , δ ) > 0. Then for all x ≥ A
ALGEBRA OF LIMITS
3
with 0 < |x − a| < δ we have
|f (x)g(x) − lm| ≤ |f (x) − l||g(x)| + |g(x) − m||l|
<
(|g(x)| + |l|)
2X
≤
(X + X)
2X
=
(above)
(as 0 < |x − a| < δ)
(choice of X and δ 0 )
as required.
(3) We show that, if m 6= 0 and f (x) 6= 0 for all n, then 1/f (x) → 1/m. To
get the full statement, then use part 2.
Let > 0. Since m 6= 0 and f (x) → m, there is δ 0 > 0 such that
|f (x) − m| < m/2 for 0 < |x − a| < δ 0 ; so for 0 < |x − a| < δ 0 , |f (x)| ≥ |m|/2
(by the triangle inequality). There also exists δ1 > 0 such that |f (x)−m| <
|m|2 /2 for all x ∈ A with 0 < |x − a| < δ1 . Let δ = min(δ1 , δ 0 ) > 0. Then,
for 0 < |x − a| < δ, we have:
1
1 m − f (x) −
=
f (x) m mf (x) |m − f (x)|
|m||f (x)|
|m|2 /2
<
|m||f (x)|
|m|2 /2
≤
|m|2 /2
= .
=
(as 0 < |x − a| < δ1 )
(as 0 < |x − a| < δ 0 )