Proving pointwise convergence of the following sequence of functions 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.
 A: 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.
A: 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
$$
e^{\alpha\log n+x-n^2x^2}\stackrel{n\to\infty}\longrightarrow 0.
$$
A: 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 $n\geq N$, 
$$
\lvert f_n(x)\rvert=n^\alpha \lvert x\rvert  e^{-n^2x^2}
=\frac{n^\alpha \lvert x\rvert}{e^{n^2x^2}}
=\frac{n^\alpha \lvert x\rvert}{\sum_{j=0}^\infty (n^2x^2)^j/j!}
\leq \frac{n^\alpha \lvert x\rvert}{n^{2k}x^{2k}/k!}
=\frac{k!\lvert x\rvert^{1-2k}}{n^{2k - \alpha}}
<\epsilon
$$
