Because of Theorem 5.47 any function that is continuous on (0, 1) but unbounded cannot be uniformly continuous there. Give an example of a continuous function on (0, 1) that is bounded, but not uniformly continuous.

Answer:[tex]f: (0,1) \to \mathbb{R}[/tex][tex]f(x) = \sin(1/x)[/tex]Step-by-step explanation:f is continuous because is the composition of two continuous functions:[tex]g(x) = \sin(x)[/tex] (it is continuous in the real numbers)[tex]h(x) = 1/x[/tex] (it is continuous in the domain (0,1))It is bounded because [tex]-1 \leq \sin(\theta) \leq 1[/tex]And it is not uniformly continuous because we can take [tex]\varepsilon = 1[/tex] in the definition. Let [tex] \delta > 0[/tex] we will prove that there exist a pair [tex]x,y\in \mathbb{R}[/tex] such that [tex]|x-y|< \delta[/tex] and [tex]|f(x) -f(y)|> \varepsilon = 1[/tex].Now, by the archimedean property we know that there exists a natural number N such that[tex] \frac{1}{N} < 2\pi \delta[/tex][tex]\Rightarrow \frac{1}{2\pi N} < \delta[/tex].Let's take [tex]x = \frac{1}{2\pi N + \pi/2}[/tex] and [tex]y = \frac{1}{2\pi N + 3\pi/2}[/tex]. We can see that [tex]|x-y| = \frac{1}{2\pi N + \pi/2}-\frac{1}{2\pi N + 3\pi/2}<\frac{1}{2\pi N} <\delta[/tex]And also:[tex]|f(x)- f(y)| = |f(2\pi N + \pi/2) - f(2\pi N + 3\pi/2)| = |1 - (-1)| = 2 > \varepsilon[/tex]And we conclude the proof.