I have to solve this: $$ \lim_{n\rightarrow\infty}\int_0^1\frac{e^{-nt}-(1-t)^n}{t}dt $$ (hint: $1-e^{nt}(1-t)^n=\int_0^t ne^{\tau n}\tau(1-\tau)^{n-1}d\tau$).
I wrote that: $$ \lim_{n\rightarrow\infty}\int_0^1\frac{e^{-nt}-(1-t)^n}{t}dt=\lim_{n\rightarrow\infty}\int_0^1\frac{1-e^{nt}(1-t)^n}{e^{nt}\,t}dt $$ and then I tried to use the hint but I don't know how to use it in a good way. I think this exercise should be resolved using the dominated convergence theorem, but I don't know how to continue.
Let $x=nt$ then $$\int_0^1\frac{e^{-nt}-(1-t)^n}{t}dt=\int_0^n \frac{e^{-x}-\left(1-\frac x n\right)^n}{x}dx=\int_0^\infty\frac{e^{-x}-\left(1-\frac x n\right)^n}{x}\chi_{(0,n)}(x) dx$$
Now we have $(1-x/n)^n\ge 1-x$ so $$\frac{e^{-x}-\left(1-\frac x n\right)^n}{x}\le 1\quad \forall x\in(0,1)$$ and we have clearly $$\frac{e^{-x}-\left(1-\frac x n\right)^n}{x}\le e^{-x}\quad \forall x\ge1$$ hence by the dominated convergence theorem we have $$\lim_{n\to\infty} \int_0^1\frac{e^{-nt}-(1-t)^n}{t}dt=0$$