For each natural number n and each number x∈[0,1] letfn(x)=xnx+1

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Jaylen Cantrell

2022-04-06

For each natural number n and each number

x \in [0, 1]

let

f_{n} (x) = \frac{x}{n x + 1}

Dallelopeelvep2yc

Beginner2022-04-07Added 15 answers

f_{n}

clearly converges to

g (x) = 0

pointwise.
Indeed,

f_{n} (0) = 0

for all n. While, for

0 < x \leq 1

, we have

\frac{1}{x} \geq 1

; whence:

| f_{n} (x) | = | \frac{x}{n x + 1} | = \frac{1}{n + \frac{1}{x}} \leq \frac{1}{n + 1} \to 0

(1)
The above actually shows that the convergence is uniform: we can make

f_{n} (x)

small for all x by taking n sufficiently large.
To be formal:
Let

ϵ > 0

. Choose N so that

\frac{1}{N + 1} < ϵ

. Then if

n \geq N

, we have, using (1):

| f_{n} (x) - 0 | = | f_{n} (x) | \leq \frac{1}{n + 1} \leq \frac{1}{N + 1} < ϵ

for all

0 < x \leq 1

Also,

| f_{n} (0) | = 0 < ϵ

. Thus, the convergence is uniform.

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