Can sin(x) Uniformly Convergence On [a, b] ?

Show that the function

\(f(x) = \sin(x)\) is uniformly continuous on \([0, \infty].\)

Sulution:

For uniformly continious on \([ 0, \infty]\) if \(\forall \varepsilon, \exists \delta > 0, \forall x, y \in [0, \infty] with\ | x - y| < \delta\) \(\Rightarrow | f(x) - f(y) | < \varepsilon\) \(we\ have\ , f(x) = \sin(x)\) \(let\ we\ choose\ \delta = \varepsilon\ then\ x, y \in [0, \infty] with\ , |x - y | < \delta\)

\[| \sin(x) - \sin(y) |\] \[\Rightarrow|2\cos\frac{x + y}{2}.\sin\frac{x - y}{2} |\] \[\Rightarrow2|\cos\frac{x + y}{2}|.|\sin\frac{x - y}{2}|\] \[Since\ |\sin(x) | \leq |x|\ for\ all\ X\] \[|\cos(x)| \leq 1\] \[\Rightarrow 2.1.\frac{|x - y|}{2}\] \[\Rightarrow |x - y| < \varepsilon\]

Hence, the function \(f(x) = \sin(x)\) is uniformly continious on \([0, \infty]\)