1
$\begingroup$

Let $f_n:(0,1)\rightarrow \mathbb{R}$ be given by $$f_n(x)=n\,x^n.$$

I want to compute

$$lim_{n \rightarrow \infty}\int^{1}_{0}f_n(x)dx.$$

I have the following:

\begin{align} lim_{n \rightarrow \infty}\int^{1}_{0}f_n(x)dx &= lim_{n\rightarrow \infty}\int_0^1 n\,x^n dx\\ &=lim_{n\rightarrow \infty}\left[\frac{n\,x^{n+1}}{n+1}\right]_{x=0}^{x=1}\\ &=lim_{n\rightarrow \infty}\left[\frac{n}{n+1}\right]\\ &=1 \end{align}

I just want to confirm this is correct.

$\endgroup$

1 Answer 1

1
$\begingroup$

Seems fine, just minor edit to remove $x=1$ and $x=0$ after you substituted them.

\begin{align} &\lim_{n\rightarrow \infty}\left[\frac{n\,x^{n+1}}{n+1}\right]_{x=0}^{x=1}\\ &=\lim_{n\rightarrow \infty}\left[\frac{n}{n+1}\right]\\ &=1 \end{align}

$\endgroup$
2
  • $\begingroup$ Actually why can’t dominated convergence theorem be applied here? $\endgroup$
    – Szeto
    Commented Nov 4, 2018 at 5:07
  • $\begingroup$ if the computation is correct, it means $f_n$ can't be dominated. Especially due to value close to $1$. $\endgroup$ Commented Nov 4, 2018 at 5:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .