Let $f_n(x)=n^\alpha x e^{-n^2x^2}$ where $\alpha \in \mathbb{R}$
Prove that $f_n$ converges pointwise to $0$.
I know that this problem has to do with polynomial growth being weaker than exponential growth, but I'm not able to do the rigorous epsilon-N proof.
On
For $x=0$, clearly $f_0(x)=0$ and so $\lim_{n\to\infty}f_n(0)=0$. Recall that $1+x\leqslant e^x$ for real $x$. Hence, \begin{align} f_n(x) &= n^\alpha x e^{-n^2x^2}\\ &\leqslant n^\alpha(1+x)e^{-n^2x^2}\\ &\leqslant n^\alpha e^xe^{-n^2x^2}\\ &=n^\alpha e^{x-n^2x^2}. \end{align} Moreover, $n^\alpha = e^{\alpha\log n}$, so $$ f_n(x) \leqslant e^{\alpha\log n+x-n^2x^2}. $$ Since $x\mapsto e^x$ is continuous, $\lim_{n\to\infty} e^{\log f_n(x)} = e^{\lim_{n\to\infty}\log f_n(x)}$. Since $x^2>0$ for $n\ne0$, we have $$ \alpha\log n + x - n^2x^2\stackrel{n\to\infty}\longrightarrow -\infty, $$ and hence $$ e^{\alpha\log n+x-n^2x^2}\stackrel{n\to\infty}\longrightarrow 0. $$
On
First fix $x\in \mathbb{R}$ and Let $k \in \mathbb{N}$ be such that $2k > \alpha$.
Now, for any $\epsilon >0$ let $N\in\mathbb{N}$ be such that $$ N^{2k-\alpha} >\frac{k!\lvert x\rvert^{1-2k}}{\epsilon} $$ Then for all $n\geq N$, $$ \lvert f_n(x)\rvert=n^\alpha \lvert x\rvert e^{-n^2x^2} =\frac{n^\alpha \lvert x\rvert}{e^{n^2x^2}} =\frac{n^\alpha \lvert x\rvert}{\sum_{j=0}^\infty (n^2x^2)^j/j!} \leq \frac{n^\alpha \lvert x\rvert}{n^{2k}x^{2k}/k!} =\frac{k!\lvert x\rvert^{1-2k}}{n^{2k - \alpha}} <\epsilon $$
The type of proof depends on what you are allowed to assume. L'Hospital's rule, for example, gives one rigorous proof (not involving explicit $\epsilon-N$ arguments).
Claim. For any $\alpha \in \mathbb{R}$ and $b > 0$, $\lim_{n \to \infty} \frac{n^\alpha}{e^{bn}} = 0$.
Proof. Apply l'Hospital's rule $\lceil\alpha\rceil$ times; then the top will be a constant, and the bottom will be $b^{\lceil\alpha\rceil}e^{bn}$, which blows up as $n$ goes to infinity, hence the whole thing vanishes in the limit.
Using this, it is easy to verify pointwise convergence; your $x$ is a fixed constant (since we are looking at pointwise convergence), and your denominator is $e^{n^2 x^2}$, which is even better (i.e. grows faster) than $e^{nx^2}$, so our claim finishes.