Suppose that we want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^1 \frac{n x^n}{1+x^3} \, dx$. Clearly the standard limit theorems for integration (Lebesgue dominated convergence theorem, monotone convergence theorem, etc.) don't work here, as the answer is $1/2$. There's at least three ways to prove this converges to $1/2$.
$1) \ $ Set $u = x^n$. This converts the problem to $\displaystyle \lim_{n \rightarrow \infty} \int_0^1 \frac{u^\frac{1}{n}}{1+u^\frac{3}{n}} \, du$ and now using the bounded convergence theorem this is trivially $1/2$
$2) \ $ Integrating by parts, setting $dv = nx^n \, dx$ and $u = \frac{1}{1+x^3}$. Then $u(1) v(1) - u(0)v(0) = \frac{n}{2(n+1)}$ while $\displaystyle \lim_{n \rightarrow \infty} \int_0^1 v \, du = 0$ rather trivially
$3) \ $ Avoiding any calculus tricks: Fix small $\epsilon > 0$ and write $\displaystyle \int_0^1 \frac{n x^n}{1+x^3} \, dx = \int_0^{1-\epsilon} \frac{n x^n}{1+x^3} \, dx + \int_{1-\epsilon}^1 \frac{n x^n}{1+x^3} \, dx$. As $n \rightarrow \infty$ the first integral converges to zero by uniform convergence. In the second one, we have $\frac12 \leq \frac{1}{1 + x^3} \leq \frac{1}{1 + (1-\epsilon)^3}$ while $\lim_{n \rightarrow \infty}\int_{1-\epsilon}^1 nx^n \, dx = 1$.
To make this precise, the easiest thing (in my opinion) is to play with $\liminf$ and $\limsup$ and then let $\epsilon \rightarrow 0^+$, which (again in my opinion) makes performing $3)$ very useful pedagogical (as is breaking an integral up and estimating each piece, instead of using calculus).
So my question is: are there similar examples as above (concrete examples where the standard limit theorems of integration don't apply) where you $\textit{can't}$ use calculus tricks to evaluate the limit, and where you $\textit{must}$ estimate like in $3)$ (or estimate in a more sophisticated manner?)