Answer:Hence shown
Step-by-step explanation:
Given
[tex]F\left ( x\right )=\frac{1}{x}[/tex]
We have to show the function is uniformly continuous in interval [2,∞)
A function is said to be continuous in given interval if it is differentiable in that interval
so derivative of [tex] \frac{1}{x}[/tex] is
[tex]\frac{\mathrm{d} \frac{1}{x}}{\mathrm{d} x}[/tex]
[tex]{F}'(x)=-\frac{1}{x^2}[/tex]
[tex]{F}'(x) is\ valid\ for\ al\l values\ of\ x\ except\ x=0[/tex]
therefore [tex] \frac{1}{x}[/tex] is continuous in [2,∞) and not uniformly continuous in (0,∞] because [tex] {F}'(x)=-\frac{1}{x^2}[/tex] is inderminate at x=0 as [tex]\frac{1}{0}[/tex] is not defined.