Help proving that these functions don't converge in $L_1$ norm

Question

so consider this

$f_n(x)= \sin(nx)$

$f=0$

$f_n$ and $f$ $\in$ $L^{1}(X)$

$X=[0,2\pi]$

Now it's true that

$ \int_E f_n$ $\rightarrow$ $\int_E f$, when $E$ contained in $X$ makes the integral of the function exist. Now why isnt it true that $\left\lVert f_n-f\right\rVert_1$ in the $L^{1}$ norm goes to zero??

Answer 1

Well, calculate $||f_n-f||_1=\int_0^{2\pi} |f_n(x)| \ dx$. Does this go to zero as $n$ goes to $\infty$?