convergence of sequence of functions on an infinite interval 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.

Hints:
$$f_n(x) = \int_0^x\sin^2\left(t+\frac1{n}\right)\,dt=\int_0^x \frac1{2}\left[1-\cos\left(2t+\frac{2}{n}\right)\right]\,dt$$
Then consider bounding $|f_n(x)-f(x)|$ using the inequality $|\sin a - \sin b|\leqslant |a-b|.$
