How to calculate this? `lim_(x->oo)1/x int_0^xdt/(2+cost)`.

Note that `int_0^oodt/(2+cost)` diverges. Without having to integrate the function, we see that `lim_(trarroo)1/(2+cost)` oscillates between values `1/(2+1)=1/3` and `1/(2-1)=1` since `cost` has a range from `-1` to `1`.

As the function is always positive and never converges to `0`, the area under the curve will also diverge, and the integral increases without bound.

Then, we see that `lim_(xrarroo)1/x int_0^xdx/(2+cost)` is really in the form `oo/oo`, where `int_0^xdt/(2+cost)` is the numerator that approaches `oo` and `x` is the denominator.

We can then apply l'Hopital's rule since we have the indeterminate form `oo/oo` by taking the derivative of the numerator and denominator separately.

`lim_(xrarroo)1/x int_0^xdt/(2+cost)=lim_(xrarroo)1/(d/dx(x))(d/dxint_0^xdt/(2+cost))`

The derivative of the integral is found through the Second Fundamental Theorem of Calculus.

`=lim_(xrarroo)1/1(1/(2+cosx))=lim_(xrarroo)1/(2+cosx)`

As we already determined, this is an oscillating function and it doesn't converge.

`1/(2+cost)` is a periodic positive function with period `T=2pi` so we have

`lim_(x->oo)1/x int_0^x (dt)/(2+cost) =1/(2pi) int_0^(2pi) (dt)/(2+cost)=1/sqrt3`

NOTE:

`lim_(x->oo)1/x int_0^x (dt)/(2+cost) = lim_(k->oo)1/(2pi k)(k int_0^(2pi) (dt)/(2+cost)) = 1/(2pi)int_0^(2pi)(dt)/(2+cost)=1/sqrt3`