Hint. If $x>0$ then there is an integer $N$ such that $1/N
P.S. Note that $$\sup_{[0,+\infty)}|g_n(x)-g(x)|=g(1/n)=1$$ where $g=0$ is the pointwise limit. Hence $g_n$ does not converge unformly to $g$ in $[0,+\infty)$.
Hint. If $x>0$ then there is an integer $N$ such that $1/N
P.S. Note that $$\sup_{[0,+\infty)}|g_n(x)-g(x)|=g(1/n)=1$$ where $g=0$ is the pointwise limit. Hence $g_n$ does not converge unformly to $g$ in $[0,+\infty)$.