# Show what is convrgent f(x)/sinx?f(x)=x-x raised to cub

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You need to check if `f(x)/sinx ` converges, hence you need to evaluate the limit of `f(x)/sinx` .

`lim_(x-gtoo) f(x)/sinx = =gtlim_(x-gtoo) (x - x^3)/sinx`

`lim_(x-gtoo) (x - x^3)/sinx = lim_(x-gtoo) (x)/sinx - lim_(x-gtoo) (x^3)/sinx`

Replacing x by `oo` yields:

`lim_(x-gtoo) (x - x^3)/sinx = oo/oo` (indeterminate)

You may use l'Hospital's rule such that:

`lim_(x-gtoo) (x - x^3)/sinx = lim_(x-gtoo) ((x)')/(sinx)' - lim_(x-gtoo) ((x^3)')/((sinx)')`

`` `lim_(x-gtoo) (x - x^3)/sinx = lim_(x-gtoo) 1/cos x - lim_(x-gtoo) 3x^2/cos x`

`` `lim_(x-gtoo) (x - x^3)/sinx = 1/oo - oo/oo`

You need to use l'Hospital's rule to evaluate `lim_(x-gtoo) 3x^2/cos x ` such that:

`lim_(x-gtoo) 3x^2/cos x =lim_(x-gtoo) ((3x^2)')/(cos x)'`

`lim_(x-gtoo) 3x^2/cos x =lim_(x-gtoo) 6x/(-sin x)`

`lim_(x-gtoo) 6x/(-sin x) = lim_(x-gtoo) ((6x)')/(-sin x)'` ``

`lim_(x-gtoo) 6x/(-sin x) = lim_(x-gtoo) 6/(-cos x) = 6/-oo = 0`

`lim_(x-gtoo) (x - x^3)/sinx = 0 - 0`

`` `lim_(x-gtoo) (x - x^3)/sinx = 0`

**The series `f(x)/sinx` converges to 0 if x goes to `oo` .**