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.
 A: I thought that it might be of interest to post a solution without appealing to the Dominated Convergence Theorem.  To that end, we now proceed.

Enforcing the substitution $t\mapsto t/n$ in the integral of interest, we find that
$$\begin{align}
\int_0^1 \frac{e^{-nt}-(1-t)^n}{t}\,dt&=\int_0^n \frac{e^{-t}-(1-t/n)^n}{t}\,dt\\\\
&=\int_0^n \frac{e^{-t}}t \left(1-e^t(1-t/n)^n\right)\,dt\tag1
\end{align}$$

Next, we have the estimates
$$\begin{align}
\left|1-e^t(1-t/n)^n\right|&\le 1-(1-t^2/n^2)^n\\\\
&\le t^2/n\tag2
\end{align}$$

Using the estimate from $(2)$ in $(1)$ reveals
$$\left|\int_0^n \frac{e^{-t}}t \left(1-e^t(1-t/n)^n\right)\,dt\right|\le \frac1n \int_0^n te^{-t}\,dt\le \frac1n$$
And we are done!

To use the Dominated Convergence Theorem, simply note that
$$\left|\frac {e^{-t}-(1-t/n)^n}{t} \xi_{[0,n]}(t)\right|\le \frac{e^{-t}-(1-t)\xi_{[0,1]}(t)}t$$, which is absolutely integrable.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\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.
