Why distributional derivative of $\frac{\sin nx}{n}\to 0$

Question

I come across following example

But I done following calculation

$<DI_{f_j},\phi>=-<I_{f_j},D\phi>=-\frac{1}{j}\int_{\mathbb R}\sin(jx)\phi'(x)dx=-\frac{1}{j}[\sin(jx)\phi(x)]_{-\infty}^{\infty}-\int_R \cos(jx)\phi(x)dx=-\int_R \cos(jx)\phi(x)dx$

So derivative must be cos(jx) I do not know Why Auther Sir Had written this?

Please Help me

Any Help will be appreciated

Answer 1

It is not the derivatives which are equal to 0, but rather the limit of the sequence of derivative. Which is true. (A quick proof not using the distribution theory consists of using the Riemann Lebesgue theorem)