Using Theorem 1, discuss lim n→∞ fn on B and C (as in Example(a)) for each of the following. (i) fn(x) = x n ; B = E1; C = [a,b] ⊂ E1. (ii) fn(x) = cos x + nx n ; B = E1. (iii) fn(x) = Xn k=1xk; B = (−1, 1); C = [−a, a], |a| < 1. (iv) fn(x) = x 1 + nx ; C= [0, +∞). [Hint: Prove that Qn = sup 1 n 1 − 1 nx + 1 = 1 n .] (v)fn(x) = cosn x; B = 0, π 2 , C = h1 4 , π 2 ; (vi) fn(x) = sin2 nx1 + nx ; B = E1. (vii) fn(x) = 1 1 + xn ; B = [0, 1); C = [0, a], 0< a < 1.

THeorem 1

Theorem 1. Given a sequence of functions fm: A → (T, ρ′), let B⊆ A and
Qm = sup
x∈B
ρ′(fm(x), f(x)).
Then fm → f (uniformly on B) iff Qm → 0.
Proof. If Qm → 0, then by definition
(∀ ε > 0) (∃ k) (∀m > k) Qm < ε.
However, Qm is an upper bound of all distances ρ′(fm(x), f(x)), x ∈B. Hence
(2) follows.
Conversely, if
(∀ x ∈ B) ρ′(fm(x), f(x)) < ε,
then
ε ≥ sup
x∈B
ρ′(fm(x), f(x)),
i.e., Qm ≤ ε. Thus (2) implies
(∀ ε > 0) (∃ k) (∀m > k) Qm ≤ ε
and Qm → 0.
Examples.
(a) We have
lim
n→∞
xn = 0 if |x| < 1 and lim
n→∞
xn = 1 if x = 1.
Thus, setting fn(x) = xn, consider B = [0, 1] and C = [0, 1).
We have fn → 0 (pointwise) on C and fn → f (pointwise) on B,with
f(x) = 0 for x ∈ C and f(1) = 1. However, the limit is not uniformon
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230 Chapter 4. Function Limits and Continuity
C, let alone on B. Indeed,
Qn = sup
x∈C |fn(x) − f(x)| = 1 for each n.2
Thus Qn does not tend to 0, and uniform convergence fails byTheorem 1.
(b) In Example (a), let D = [0, a], 0 < a < 1. Then fn → f(uniformly) on
D because, in this case,
Qn = sup
x∈D |fn(x) − f(x)| = sup
x∈D |xn − 0| = an → 0.
(c) Let
fn(x) = x2 +
sin nx
n
, x ∈ E1.
For a fixed x,
lim
n→∞
fn(x) = x2 since

sin nx
n

≤
1
n → 0.
Thus, setting f(x) = x2, we have fn → f (pointwise) on E1.Also,
|fn(x) − f(x)| =

sin nx
n

≤
1
n
.
Thus (∀ n) Qn ≤ 1
n → 0. By Theorem 1, the limit is uniform on all of
E1.
Note 1. Example (a) shows that the pointwise limit of a sequence ofcon-
tinuous functions need not be continuous. Not so for uniformlimits, as the
following theorem shows.