Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$? I am trying to show that
$$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$
I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty}  (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$
I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$
 A: Let
$f_n(x)
=e^{-nx} - (1-x)^n
$.
I will show that
for $0 < x < 1$,
$f_n(x) \ge 0$
and
if $x > \dfrac1{\sqrt{n}}$
then
$f_n(x)
\lt e^{-\sqrt{n}}
$.
and if
$0 < x < c_0
\approx 0.683803
$
(the root of
$\ln(1-x)+x+x^2
=0$)
then
$f_n(x) < x
$.
Therefore
$0 < f_n(x)
\le\dfrac1{\sqrt{n}}
$.
For $0 < x < 1$,
$e^x < \dfrac1{1-x}
$
(compare the power series)
so
$e^{-x} > 1-x
$
so
$e^{-nx} > (1-x)^n
$.
Therefore
$f_n(x) > 0$.
If $x > \dfrac1{\sqrt{n}}$ then
$e^{-nx}
\lt e^{-\sqrt{n}} 
$
so
$f_n(x)
\lt e^{-\sqrt{n}}
$.
$\begin{array}\\
f_n(x)
&=e^{-nx}-(1-x)^n\\
&=e^{-nx}-e^{n\ln(1-x)}\\
&=e^{-nx}(1-e^{nx+n\ln(1-x)})\\
&<e^{-nx}(1-e^{nx+n(-x-x^2)})
\qquad 0 < x < c_0\ (*)\\
&=e^{-nx}(1-e^{-nx^2})\\
&<e^{-nx}(1-(1-nx^2))\\
&=nx^2e^{-nx}\\
&\le x
\qquad\text{since } z e^{-z} \le 1\\
\end{array}
$
$(*)
\ \ln(1-x)
\gt -x-x^2
$
for
$0 < x < c_0
\approx 0.683803
$.
A: Use the result:
$$\color{red}{0\le e^{-nx}-(1-x)^n\le nx^2 e^{-nx} \qquad (0\le x\le 1),}$$
which can be refered in key inequality.
When $n>2$, $f_n(x)=nx^2 e^{-nx}$ has maximum at $\frac{2}{n}\in(0,1)$.
So
$$\sup_{x\in[0,1]}\left(e^{-nx}-(1-x)^n\right)\leq\frac{4}{n}e^{-2}\to 0.$$
A: To get rid of the $e^{-n x}$ term, use the formula that you just found: $\frac{n}{1-n} = \frac{\log(1-x)}{x}$ (just move the $x$ in the denominator of the RHS to the LHS). That is going to give you a formula of the form $(1-x)^{n-1} - (1-x)^n$. From here, the result should be almost immediately.
A: actually if you focus on the limit
$$\lim_{n \to \infty}\displaystyle e^{\displaystyle-nc} -(1-c)^n$$
where c is a real number between $0$ and $1$then if $c \neq 0$ then the exponential term goes to zero and the other bracket also and if $c = 0$ the value of the limit is zero so this can be proved directly without evaluating the maximum
