Respuesta :

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.

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