Let $f_n(x) = n^\alpha x^n(1-x)$ for any fixed $\alpha \in \mathbb{R}$ and $x \in [0,1]$. Prove $f_n$ converges to $0$ pointwise (via the definition).
I considered that
$|n^\alpha x^n(1-x) - 0| = n^\alpha x^n (1-x) \leq n^\alpha x^n$.
But then I'm stuck here, because it seems that I can't isolate the $n$ term, to get a relationship between $n$ and $\epsilon, x$.
The cases $x=0$ and $x=1$ are trivial, so we can limit ourselves to $x \in (0,1)$. The factor $1-x$ is simply a constant and is irrelevant to convergence. As such, all we need is $n^{\alpha} x^n$ to go to zero as $n$ goes to infinity. A 'trick' for this could be to use the ratio test on $\sum n^\alpha x^n$:
$$\dfrac{(n+1)^\alpha x^{n+1}}{n^\alpha x^n} = \left(\dfrac{n+1}{n}\right)^\alpha \cdot x.$$
Note how $\lim_{n\to\infty} \frac{n+1}{n} = 1$, so because $|x| < 1$, we have: $$ \limsup_{n\to\infty} \left | \dfrac{(n+1)^\alpha x^{n+1}}{n^\alpha x^n} \right | < 1,$$ so the series $\sum n^\alpha x^n$ converges. This means its individual terms go to zero, so we finally have $n^\alpha x^n \to 0$ as $n \to \infty$.