$f_n^\alpha(x) = n^\alpha x^n$ converges almost everywhere

Question

In [0,1], with the Lebesgue measure and for $\alpha \in \mathbb{R}$, I must show that $f_n^\alpha(x):=n^\alpha x^n$ converges almost everywhere in [0,1] and identify the limit.

It is clear to me that I've to build some set $E \subset [0,1]$ such that $f_n^\alpha$ converges in E and $|E^c|=0$, but I am struggling to see the limit and the construction of such a set.

Coud somone give me a hand?

Answer 1

Rewrite it as (for $x\in(0,1)$) $$ f_n^\alpha(x) = e^{\alpha \log(n)}\cdot e^{n\log(x)} = e^{\alpha \log(n) + n\log(x)} $$ Now for $x\in(0,1)$, $\log(x) = -\varepsilon$ for some $\varepsilon >0$. And the limit $$ \lim_{n\to\infty} \big(\alpha\log(n) -\varepsilon n\big) = -\infty $$ So $f_n^\alpha(x) \to 0$ for $x\in(0,1)$.
Now when $x=0$ you get that $$f_n^\alpha(0) = n^\alpha \cdot 0 = 0$$

Lastly when $x=1$ you get that $$f_n^\alpha(1) = n^\alpha\to\infty$$ meaning it diverges only when $x=1$.

Hence $f_n^\alpha$ converge almost everywhere.