Find $\lim\limits_{n \to \infty} n \int_2^e (\ln x)^n \, dx$. I have to find the limit:
$$\lim\limits_{n \to \infty} n \displaystyle\int_2^e (\ln x)^n dx$$
I tried using the Squeeze Theorem but it didn't work (or at least I didn't use it correctly):
$$2 \le x \le e$$
I took the log of this inequality
$$\ln(2) \le \ln(x) \le 1$$
Raised to the $n$:
$$\ln^n(2) \le \ln^n(x) \le 1$$
Took the integral from $2$ to $e$:
$$\int_2^e \ln^n(2)dx \le \int_2^e \ln^n(x) dx \le \int_2^e 1 dx$$
$$\ln^n(2) \cdot (e-2) \le \int_2^e \ln^n(x) dx \le (e-2)$$
Multiplied by n:
$$n \ln^n(2) \cdot(e - 2) \le n\int_2^e \ln^n(x) dx \le n(e-2)$$
And now, since $\ln(2) < 1$, so $\ln^n(2) = 0$ as $n \to \infty$, if I take the limit with $n \to \infty$ I get:
$$\infty \cdot 0 \le n\int_2^e \ln^n(x) dx \le \infty$$
So I didn't get anywhere trying to use the Squeeze Theorem.
 A: As you said $\lim\limits_{n\to \infty}\ln^n 2=0$. Let:
$$I_n = \int_2^e\ln^n x\,dx$$
$I_n$ is decreasing because:
$$I_{n}-I_{n+1}=\int_2^e\ln^n x(1-\ln x)\,dx \geq 0$$
Integrating by parts:
$$
\begin{aligned} 
\displaystyle\int_2^e \ln^n x\, dx &= \displaystyle\int_2^e \ln^n x \cdot (x)'\, dx\\
&=x\ln^n x\bigg|_2^e -n\int_2^e\ln^{n-1}x\,dx\\
&= e-2\ln^n 2-nI_{n-1}
\end{aligned}
$$
Therefore $I_n+nI_{n-1}=e-2\ln^n2$. Now, since $I_n$ is decreasing:
$$I_n+nI_{n-1}\geq I_n+nI_n\Rightarrow I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$
and
$$I_{n+1}+(n+1)I_n\leq I_n+(n+1)I_n\Rightarrow I_n\geq \frac{1}{n+2}(e-2\ln^{n+1}2)$$ 
Combining the two:
$$\frac{1}{n+2}(e-2\ln^{n+1}2) \leq I_n\leq \frac{1}{n+1}(e-2\ln^n 2)$$
or
$$\frac{n}{n+2}(e-2\ln^{n+1}2) \leq nI_n\leq \frac{n}{n+1}(e-2\ln^n 2)$$
and squeezing $\lim\limits_{n\to \infty}nI_n=e$. 
A: Use $t = \log^{n+1}(x) \implies \frac{1}{n+1}e^{t^{\frac{1}{n+1}}}dt = \log^n(x)dx$:
$$I = \lim_{n\to\infty} \frac{n}{n+1}\int_{\log^{n+1}(2)}^1 e^{t^{\frac{1}{n+1}}}\:dt \longrightarrow \int_0^1 e\:dt = e$$
by dominated convergence.
A: Let $\epsilon $ be any fixed number such that $0<\epsilon <e-2$ and we split the integral as a sum of two integrals using intervals $[2,e-\epsilon]$ and $[e-\epsilon, e] $. The first integral tends to $0$ as integrand uniformly tends to $0$.
For the second integral we need to bound it suitably. We have via integration by parts $$(n+1)\int_{e-\epsilon} ^{e} \dfrac{(\log x) ^n} {x} \cdot x\, dx=\left. x(\log x) ^{n+1}\right|_{x=e-\epsilon} ^{x=e} - \int_{e-\epsilon} ^{e}(\log x) ^{n+1}\,dx$$ (note that the factor $n$ before integral can be safely replaced by $(n+1)$ as $n/(n+1)\to 1$). The first term on right tends to $e$ and second term is bounded in absolute value by $\epsilon$ (as integrand lies in $[0,1]$). It follows that the desired limit is $e$. 
A: Fixed $\epsilon\in(0,1-\ln2)$. Let $u=\ln x$ and then
$$ n\int_2^e (\ln x)^n dx=n \int_{\ln 2}^1 u^ne^u du=n \int_{\ln 2}^{1-\epsilon} u^ne^u du+n \int^1_{1-\epsilon} u^ne^u du.$$
For the 1st part,
$$ 0\le n \int_{\ln 2}^{1-\epsilon} u^ne^u du\le 3n(1-\epsilon)^n $$
and one has
$$ \lim_{n\to\infty} n \int_{\ln 2}^{1-\epsilon} u^ne^u du=0. $$
For the 1st part, by the Mean Value Theorem for integrals, there is $\xi\in(1-\epsilon,1)$ such that
$$ n \int_{1-\epsilon}^1 u^ne^u du=n e^\xi\int_{1-\epsilon}^1 u^n du=\frac{n}{n+1}e^\xi(1-(1-\epsilon)^{n+1}). $$
Noting that
$$ e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le e^\xi(1-(1-\epsilon)^{n+1})\le e  $$
one has
$$ \frac{n}{n+1}e^{1-\epsilon}(1-(1-\epsilon)^{n+1})\le n \int_{1-\epsilon}^1 u^ne^u du=n e^\xi\int_{1-\epsilon}^1 u^n du\le\frac{n}{n+1}e $$
and hence
$$  e^{1-\epsilon}\le \underline{\lim}_{n\to\infty}n \int_{1-\epsilon}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{1-\epsilon}^1 u^ne^u du\le e. $$
So
$$  e^{1-\epsilon}\le \underline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le e. $$
Letting $\epsilon\to0$ one has
$$ e\le \underline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le \overline{\lim}_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du\le e $$
or
$$ \lim_{n\to\infty}n \int_{\ln 2}^1 u^ne^u du= e $$
