Calculate lim F(x) given F(x)=integral (1<x)f (t)dt  f(x)=1/(2-sin x)

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You need to use the following definition of sine function, such that:

`sin x < 1 => -sin x < 1`

Adding 2 both sides of inequality, yields:

`2 - sin x < 2 + 1 => 2 - sin x < 3 => 1/(2 - sin x) > 1/3 => f(x) > 1/3`

Integrating both sides the inequality `1/(2 - sin x) > 1/3` yields:

`int 1/(2 - sin x) dx > int (dx)/3 = x/3`

Evaluating the limit yields:

`lim_(x->oo) int 1/(2 - sin x) dx > lim_(x->oo) x/3 = oo`

Hence, evaluating the limit `lim_(x->oo) F(x) = lim_(x->oo) int 1/(2 - sin x) dx = oo` .

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