Show that the Laplace transform is right continuous at $0$ Question:
Show using the Lebesgue dominated convergence theorem that $F(s):=\int_0^{\infty} e^{-st}f(t) \ dt$ is right continuous at $0$. Note that $f \in L^1[0,\infty]$
my work:
$F(x)$ is right continuous at $0$ if for every sequence $x_n \to 0$ (from the right), $F(x_n) \to F(0)$. It is sufficient to check the sequence $\frac{1}{n}$. In this case, $|e^{-(1/n)t}f(t)| \leq f(t)\in L^1[0,\infty]$ so the LDCT tells us that $\lim_{n \to \infty} F(1/n)=F(0)$ (that is, we can switch the limits).


It is sufficient to check the sequence $\frac1n$ 

I see no reason why checking this one sequence is sufficient. Even if such a reason can be given, you gain nothing by replacing $x_n$ with $\frac1n$ in your argument. 
Just write
$$|e^{-x_nt}f(t)| \leq| f(t)|\in L^1[0,\infty]$$
and conclude that $F(x_n)\to F(0)$ by the DCT.
