# prove that there is antiderivative for f(x)=sin(1/x),x is not 0, f(x)=0, x=0 im confused because f is not continuos in x=0

sciencesolve | Certified Educator

You should notice the discontinuity of the function at x=0 and you need to prove that the antiderivative of the function exists.

Differentiating the product `x^2*cos(1/x)`  yields:

`(x^2*cos(1/x))' = 2xcos(1/x) - x^2*sin(1/x)*(-1/x^2)`

`(x^2*cos(1/x))' = 2xcos(1/x) + sin (1/x) `

Hence, `sin (1/x) = (x^2*cos(1/x))' - 2xcos(1/x)`

You need to remember that `lim_(x-gt0) 2xcos(1/x) = 0` , hence,  the branched function g(x) =  `{(2xcos(1/x) if x!=0),(0 if x=0):}`  exists and it is continuous, hence, the function `G(x)`  is the antiderivative of `g(x)`  and `G'(0) = g(0).`

Notice  that `lim_(x-gt0) x^2cos(1/x) = 0` , hence, the function `{(x^2cos(1/x) if x!=0),(0 if x=0):}`  exists and it is continuous.

Hence, for `x!=0` , `f(x)=sin(1/x)=h'(x)-G'(x) = (h-G)'(x)` , for `x=0, f(0)= (h-G)'(0)` , hence, `h(x)-G(x)`  represents the antiderivative of the function `f(x).`