The limit of $nx^n$ Let $f_n(x) = nx^n$ for x belongs to $[0,1]$ and n belongs to natural numbers. How to show that $\lim f_n(x)=0$ for x belongs to $[0,1)$? And also, how to find the limit of $\int_0^1 f_n(x)dx$ when $n\to\infty$? 
 A: If $x \in (0,1)$ then $x = (1 + a)^{-1}$ with $a > 0.$
Hence, by the binomial theorem
$$0 \leqslant nx^n = \frac{n}{(1+a)^n}< \frac{2n}{n(n-1)a^2},$$
and $nx^n \to 0$ as $n \to \infty$ by the squeeze theorem.
Also
$$\int_0^1 f_n(x) \, dx = \int_0^1 nx^n \, dx = \left. \frac{nx^{n+1}}{n+1}\right|_0^1= \frac{n}{n+1} \to 1.$$
The sequence does not converge uniformly and you can't justify switching the limit and the integral.
As we see,
$$1 = \lim_{n \to \infty} \int_0^1f_n(x) \, dx \neq \int_0^1 \lim_{n \to \infty}f_n(x) \, dx = 0$$
A: For each $n$,
for any $c > 0$,
we need to find a
$x \in [0, 1)$
such that
$f(x)
=nx^n
< c
$.
If $x = 1-y$,
$\begin{array}\\
nx^n
&=n(1-y)^n\\
&=ne^{n\ln(1-y)}\\
&=ne^{-n(y+y^2/2+...)}\\
&<ne^{-ny}\\
&=e^{-ny+\ln(n)}\\
\end{array}
$
If 
$y \ge \frac1{\sqrt{n}}$,
then
$ny \ge \sqrt{n}
$
so
$nx^n
\le e^{-\sqrt{n}+\ln(n)}
$.
Since
$\frac{\ln(n)}{\sqrt{n}}
\to 0$
as
$n \to \infty$,
$n (1-\frac1{\sqrt{n}})^n
\to 0$.
Note that works
just as well for
$n^kx^n$
for any fixed $k$,
since this shows that
$n^k(1-\frac1{\sqrt{n}})^n
\le e^{-\sqrt{n}+k\ln(n)}
$.
