$\lim \int_1^2 \ln^n x dx$ 
a) show that $\lim \int_1^2 \ln^n x dx$ goes to $0$ as $n$ goes to $\infty$
b) show that $\lim \int_2^3 \ln^n x dx$ goes to $\infty$ as $n$ goes to $\infty$

a) on $1<x<2$, $\ln(x) < 1$, so $ln^n(x)$ goes to $0$
b) on $2<x<e$; $\ln(x) < 1$ but in $e\le x \le 3$, $\ln (x) \ge 1$ so
$\lim \int_2^3 \ln^n x dx = \lim (\int_2^e \ln^n x dx + \int_e^3 \ln^n x dx)$
$|\int_2^e \ln^n x dx + \int_e^3 \ln^n x dx|\le |\int_2^e \ln^n x dx| + |\int_e^3 \ln^n x dx|$
Now the problem for me is that  for all $\epsilon > 0$ $|\int_2^e \ln^n x dx| < \epsilon$, and for all $M > 0$, $|\int_e^3 \ln^n x dx| > M$.
What should I do next? (Computing integrals is not allowed)
 A: Let's prove something slightly stronger. Assume $a \in \mathbb{R}$, that $f: [a, \infty) \rightarrow [0, \infty)$ is strictly increasing and that $b>a$ is such that $f(b)=1$. As it is monotonous, $f$ will be Riemann-integrable on any compact interval included in its domain of definition.
It is then the case that:

On the subunitary side:
$$\int_{a}^{b} f^u (x)\mathrm{d}x \xrightarrow{u \to \infty} 0$$

Proof : Consider arbitrary $\epsilon \in (0, b-a)$. Then we have the relations:
\begin{align*}
\int_{a}^{b} f^u(x)\mathrm{d}x&=\int_{a}^{b-\frac{\epsilon}{2}}f^u(x)\mathrm{d}x+\int_{b-\frac{\epsilon}{2}}^{b}f^u(x)\mathrm{d}x \\
&\leqslant \int_{a}^{b-\frac{\epsilon}{2}}f^u\left(b-\frac{\epsilon}{2}\right)\mathrm{d}x+\int_{b-\frac{\epsilon}{2}}^{b}1\mathrm{d}x \\
&\leqslant  f^u\left(b-\frac{\epsilon}{2}\right) \cdot (b-a)+\frac{\epsilon}{2}.
\end{align*}
As $f\left(b-\frac{\epsilon}{2}\right)<f(b)=1$, there exists a $t>0$ such that for all $u>t$, one has $f^u\left(b-\frac{\epsilon}{2}\right)<\frac{\epsilon}{2(b-a)}$. Hence, for all $u>t$ one obtains from the above the estimate $0 \leqslant \int_{a}^{b} f^u(x)\mathrm{d}x < \epsilon$, q.e.d. $\Box$



*For any $x>b$ one has
$$\int_{b}^{x} f^u(t)\mathrm{d}t \xrightarrow{u \to \infty} \infty$$

Proof : perfectly analogous to the one above, you can 'majorise' by the powers of any $f(c)>1$ for arbitrary (but fixed!) $c \in (b, x)$. $\Box$
In your particular case, the role of $f$ is played by the natural logarithm, restricted to the domain $[1, \infty)$ (in other words $a$ corresponds to $1$ and $b$ to $\mathrm{e}$, as I hope you will gather).
