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 $$[0, \infty]$$

Tags:

Categories:

Updated: