# 1/x and sin(1/x) are not Uniformly Continuous on (0,1)

In this post, we will prove that the functions 1/x and sin(1/x) defined on (0, 1) are not uniformly continuous on (0, 1).

## 1/x is not Uniformly Continuous

If $f(x) = \dfrac{1}{x}$ is uniformly continuous on (0, 1) then for every Cauchy sequence {xn} on (0, 1), the sequence {f(xn)} will also be a Cauchy sequence.

Consider the sequence {xn} where

xn = $\dfrac{1}{n+1}, n \in \mathbb{N}$.

This is a Cauchy sequence in (0, 1). Now,

{f(xn)} = {2, 3, 4, …}.

This sequence {f(xn)} is not a Cauchy sequence. This shows that f does not send a Cauchy sequence to a Cauchy sequence. Hence, f(x) = 1/x defined on (0, 1) is not uniformly continuous.

We can use the definition of uniform continuity in order to show 1/x is not uniform continuous on (0, 1).

Choose any δ>0.

Then by the Archimedean Property of Real Numbers, there exists an integer m such that mδ>1. So we have that

$\dfrac{1}{n} < \delta ~\forall ~n \geq m.$

Take x1 = $\dfrac{1}{m}$ and x2 = $\dfrac{1}{2m}$ in (0, 1). Note that

|x1 – x2| = $\dfrac{1}{2m}$ < δ.

But,

|f(x1) – f(x2)| = m $\nless$ ε chosen arbitrarily.

So by definition we conclude that the function f(x) = 1/x is not uniform continuous on (0, 1).

log(x) is uniformly continuous

x2 is Uniformly Continuous on (0, 1) but NOT on (0, ∞)

## Sin(1/x) is not Uniformly Continuous

Assume $f(x) = \sin(\dfrac{1}{x})$ is uniformly continuous on (0, 1). Then for each Cauchy sequence {xn} on (0, 1), the sequence {f(xn)} will be Cauchy.

Let us consider the Cauchy sequence {xn} in (0, 1) with xn = $\dfrac{2}{n \pi}$. Then the sequence {f(xn)} becomes

{1, 0, -1, 0, 1, 0, …}

which is not Cauchy. Hence the function sin(1/x) is not uniformly continuous on (0, 1).

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## FAQs

Q1: Is 1/x uniformly continuous on (0, 1)?

Answer: No, f(x) = 1/x is not uniformly continuous on (0, 1). Because, {$\frac{1}{n+1}$} is Cauchy on (0, 1) but {$f(\frac{1}{n+1})$} = {2, 3, 4, …} is not Cauchy.

Q2: Is sin(1/x) uniformly continuous on (0, 1)?

Answer: No, sin(1/x) is not uniformly continuous on (0, 1). Because, {$\frac{2}{n \pi}$} is Cauchy on (0, 1) but {$f(\frac{2}{n \pi})$} = {1, 0, -1, 0, 1, 0, …} is not Cauchy.

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