I've recently discovered DCT and I'm wondering how one would solve this limit : The limit: $$ \lim_{n \to \infty}  n\int_{2}^{e} [\ln(x)]^n \mathrm dx $$
The book from where I took this exercise offered these as possible results :

*

*$e$

*$0$

*$1$

*$\ln(2)$

*infinity

I was able to pinpoint the solution ($e$) by substituting $x=e^t$ and then building the integral from  $\ln(2)\le t\le1$, and in the end reaching to $2\le n\int_{2}^{e} [\ln(x)]^n \mathrm dx \le e$ . The only solution possible, considering the answers the book gave, was $e$.
The problem is, I still didn't solve the problem. I am still not sure how to reach the correct result so in my attempt to find a way, I discovered DCT. I tried to understand as much as I can but there are still a lot of empty gaps.
In order to use DCT, I tried bringing the $n$ inside the integral and then use the substitution $t=(\ln(x))^n$ which gives us $ \lim_{n \to \infty}  \int_{(\ln(2))^n}^{1} t^\frac{1}{n} \mathrm dt $. Because the lower bound depends on $n$, I basically got stuck.
So with all of this said, is there a way to use DCT to solve this limit?
 A: $\newcommand{\d}{\mathrm{d}}$Too long for a comment. David's link gives you a good solution, I am just showing you your attempt would have also worked had you pursued it.
Let: $$I_n:=\int_2^e\ln^n(x)\,\d x$$Let, as you suggested, $t=\ln^n(x)$. Then $t^{1/n}=\ln x$, $x=\exp(t^{1/n})$. Then: $$\frac{\d x}{\d t}=\exp(t^{1/n})\cdot\frac{1}{n}t^{-1+1/n}$$And we have: $$I_n=\frac{1}{n}\int_{\ln^n2}^1t\cdot\exp(t^{1/n})t^{-1+1/n}\,\d t=\frac{1}{n}\int_{\ln^n2}^1t^{1/n}\exp(t^{1/n})\,\d t$$You can apply the dominated convergence theorem by observing that:

*

*$|\ln2|\lt1\therefore\ln^n2\to0$ so integrating over $(0,1]$ will provide domination

*$\lim_{n\to\infty}t^{1/n}=1$ for all $t\gt0$ and on the interval $(0,1]$ this is a monotone increasing limit

*The function $t^{1/n}\exp(t^{1/n})$ is an increasing positive function

Then the integrand $t^{1/n}\exp(t^{1/n})\chi_{[\ln^n2,1]}$ is dominated by (and converges pointwise to) $1\exp(1)\chi_{(0,1]}=e\cdot\chi_{(0,1]}$, and you can finish from here. In fact this convergence is monotonic, so the monotone convergence theorem suffices.
