Uniform Continuity of f(x)=1/x on [1, infinity) and (0, 1]

Published on June 4, 2026

Calculus

Prove that the function f(x)=1/x is uniformly continuous on [1, infinity) but not on (0, 1] using the definition of uniform continuity and sequential criterion.

Question

please answer

Answer

To prove the given statement, we need to address two parts: showing that

is uniformly continuous on

and showing that it is not uniformly continuous on

Part 1: Uniform continuity on

Definition: A function

is uniformly continuous on a set

if for every

, there exists a

such that for all

, if

, then

Proof:
Let

be given. We want to examine the difference

for

Since

and

, it follows that

. Therefore,

.
Substituting this into our inequality, we get:

To ensure

, we can choose

.
Then, for any

, if

, we have:

Thus,

is uniformly continuous on

Part 2: Not uniformly continuous on

Definition of Negation: A function

is not uniformly continuous on

if there exists some

such that for every

, there exist

with

but

Proof:
We can use the sequential criterion for non-uniform continuity: if there exist two sequences

and

such that

, but

does not converge to

, then

is not uniformly continuous.

Let’s choose the sequences:

Both

and

are in the interval

for all

Check the distance between sequences:

As , . So, .
Check the distance between function values:

As , , which is not .

Since we found two sequences whose distance approaches zero but whose images’ distance remains constant (specifically,

), the function

is not uniformly continuous on

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