The book "Berkeley Problems in Mathematics" contains some hardcore math problems which Berkeley had used for its Phd program clearing requirement in the 70s and 80s.
Here is a nice one from there.
Suppose $f:[0,1] \to \mathbb{R}$ is a continuous function. Find
$$\lim_{n \to \infty} n \int_{0}^{1} x^n f(x) dx$$
Scroll down for a couple of solutions:
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$\textbf{Solution 1}:$
This is a "heavyweight" and uses the Weierstrass approximation theorem.
The approximation theorem says that any continuous function on a closed interval can be approximated as close as we want, using just polynomials.
We first begin with the observation that for any fixed $M$
$$\lim_{n \to \infty} n \int_{0}^{1} x^n x^M dx = \lim_{n \to \infty} \frac{n}{n + M + 1} = 1$$
This implies that, for any polynomial $P$ we have that
$$\lim_{n \to \infty} n \int_{0}^{1} x^n P(x)dx = P(1)$$
For ease of notation, let $$I(n,g) = n\int_{0}^{1} x^n g(x) dx$$
Now given our $f$ and an $\epsilon > 0 $, there is a polynomial $P_{\epsilon}$ such that $$|f(x) - P_{\epsilon}(x)| < \epsilon \quad \forall x \in [0,1]$$
i,e
$$-\epsilon + P_{\epsilon}(x) \le f(x) \le \epsilon + P_{\epsilon}(x)$$
And thus we have
$$ I(n, -\epsilon) + I(n, P_{\epsilon}) \leq I(n,f) \leq I(n, \epsilon) + I(n, P_{\epsilon})$$
i.e.
$$ \frac{-n\epsilon}{n+1} + I(n, P_{\epsilon}) \leq I(n,f) \leq \frac{n\epsilon}{n+1} + I(n, P_{\epsilon})$$
Taking $n \to \infty$ we get
$$-\epsilon + P(1) \leq \liminf I(n,f) \leq \limsup I(n,f) \leq \epsilon + P(1)$$
And thus
$$-2\epsilon + f(1) \leq \liminf I(n,f) \leq \limsup I(n,f) \leq 2\epsilon + f(1)$$
$\textbf{Solution2:}$
This is more elementary.
Let $$c_n = \frac{1}{\sqrt[n]{n}}$$
Note that $c_n < 1$, $\lim c_n = 1$ and $\lim c_n^{n+1} = 0$
Now since $f$ is continuous, we can bound $f(x)$ in some interval $[m,M]$ and thus
$$\frac{mn}{n+1} c_n^{n+1} \leq n\int_{0}^{c_n} x^n f(x)dx \leq \frac{Mn}{n+1}c_n^{n+1}$$
Thus $$\lim_{n \to \infty} n \int_{0}^{c_n} x^n f(x)dx = 0$$
Since $c_n \to 1$, given an $\epsilon > 0 $, for sufficiently large $n$, we have that
$$|f(x) - f(1)| < \epsilon \quad \forall x \in [c_n, 1]$$
And we get
$$ \frac{n(f(1) - \epsilon)}{n+1} (1 - c_n^{n+1}) \leq n\int_{c_n}^{1} x^n f(x) dx \leq \frac{n(f(1) + \epsilon)}{n+1}(1 - c_n^{n+1})$$
Taking $n \to \infty$ we get
$$f(1) - \epsilon \leq \liminf n \int_{c_n}^{1} x^n f(x)dx \leq \limsup n \int_{c_n}^{1} x^n f(x)dx \leq f(1) + \epsilon$$
Since $\epsilon$ was arbitrary it follows that
$$\lim_{n \to \infty} n \int_{c_n}^{1} x^n f(x) dx = f(1)$$
Thus
$$\lim_{n \to \infty} n \int_{0}^{1} x^n f(x) dx = $$
$$\lim_{n \to \infty} n\int_{0}^{c_n} x^n f(x) dx + \lim_{n \to \infty} n\int_{c_n}^{1} x^n f(x)$$
$$ = 0 + f(1) = f(1)$$
