How to prove the limit of this integral? How to prove the limit for fucntion $f = 0 $ by Lebesgue Dominated Convergence Theorem:
$$f = \lim_{n\to \infty}\int_0^1 \frac{e^{-nt}-(1-t)^n}{t}\, dt=0$$
I believe I can use the equality: $1-e^{-nt}(1-t)^n=\int_0^t {ne^{-n\tau}\tau(1-\tau)^{n-1}}\, dt$; but  after I do this equality n times, I find I get
$$\lim_{n\to \infty}\int_0^1 \frac{e^{-nt}}{t} dt$$ However, this will explode when t = 0 but I do not know how to find the function g $\geq \lvert f\rvert$ that to apply the Lebesgue Dominated Convergence Theorem to switch the $\lim$ and $\int$ position.

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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\int_{0}^{1}
{\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
\int_{0}^{1}{\expo{-nt} - 1 \over t}\,\dd t +
\int_{0}^{1}{1 - \pars{1 - t}^{n} \over t}\,\dd t}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{%
-\int_{0}^{n}{1 - \expo{-t} \over t}\,\dd t +
\int_{0}^{1}{1 - t^{n} \over 1 - t}\,\dd t}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{-\operatorname{Ein}\pars{n} +
H_{n}}
\end{align}
$\ds{\operatorname{Ein}}$ is the
Complementary Exponential Integral And $\ds{H_{z}}$ is a
Harmonic Number.
$$
\mbox{As}\ n \to \infty,\quad
\left\{\begin{array}{lcll}
\ds{\operatorname{Ein}\pars{n}} & \ds{\sim} &
\ds{\ln\pars{n} + \gamma + {\expo{-n} \over n}} &
\ds{\color{red}{\large\S}}
\\
\ds{H_{n}} & \ds{\sim} &
\ds{\ln\pars{n} + \gamma + {1 \over 2n}} &
\ds{\color{blue}{\large\#}}
\end{array}\right.
$$
$$
\mbox{such that}\quad\bbox[5px,#ffd]{\int_{0}^{1}
{\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t} \sim
{1 \over 2n}\quad\mbox{as}\quad n \to \infty
$$
\begin{align}
& \mbox{}
\\ &\ \implies
\bbx{\bbox[5px,#ffd]{\lim_{n \to \infty}\int_{0}^{1}
{\expo{-nt} - \pars{1 - t}^{n} \over t}\,\dd t} = 0} \\ &
\end{align}

$\ds{\color{red}{\large\S}}$:
See this link and this one.
$\ds{\color{blue}{\large\#}}$: The asymptotic $\ds{H_{z}}$ behavior is given in the
above cited link.
