# 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

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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).` **