For $n\ge 1$, Let $$g_n(x)=\sin^2(x+\frac{1}{n}), x\in[0,\infty)$$ and $$f_n(x)=\ \int_{0}^{x}g_n(t)dt$$. Then
1) $\{f_n\}$ converges pointwise to a funtion $f$ on $[0,\infty)$ but does not converge uniformly on $[0,\infty)$
2) $\{f_n\}$ does not converge pointwise to any function on $[0,\infty)$
3) $\{f_n\}$ converges uniformly on $[0,1]$
4) $\{f_n\}$ converges uniformly on $[0,\infty)$.
i found the pointwise limit of $g_n(x)$ then I don't know how to solve.