Show (pointwise) convergence of function sequence $f_n :[0,1] \to \mathbb{R}$ with $f_n(t)=nxe^{-nx^2}$ Show (pointwise) convergence of function sequence $f_n :[0,1] \to \mathbb{R}$ with $f_n(x)=nxe^{-nx^2}$
My attempt:
Let $\varepsilon >0$, then there exists $N\in \mathbb{N}$ such that $N>e^x\cdot \varepsilon$; such that  $n>N$ and $x\in [0,1]$ with:$$
\left | n\cdot e^{-nx^2}\right |=n\cdot  \left |e^{-nx^2}\right|\leq n\cdot  e^{-x} <\varepsilon$$
Is that correct?
 A: If $x=0$ then $f_n(x)=0$ and trivially $\lim_{n\to\infty}f_n(x)=0$ For $x\in (0,1]$, we have $f_n(x) = nxe^{-nx^2}\leqslant ne^{-nx^2}=ne^{-n}e^{-x^2}$. Note that $x\mapsto e^{-x^2}$ is continuous and decreasing on $(0,1]$, so $f_n(x)\leqslant ne^{-n}\stackrel{n\to\infty}\longrightarrow 0$.
In fact, $f_n$ converges uniformly on $[0,1]$. Consider
$$
\frac{\mathsf d}{\mathsf dx} f_n(x) = e^{-nx^2}(n-2n^2x^2).
$$
Since $e^{-nx^2}>0$ for all $x$, it is clear that the derivative is zero only when $n-2n^2x^2=0$, or $x = \frac1{\sqrt{2n}}$ (we choose the positive root because we are considering $x\in[0,1]$. Let $x_n = \frac1{\sqrt{2n}}$, then
$$
\sup_{x\in[0,1]} |f_n(x)|\leqslant f_n(x_n) =  \frac n{\sqrt{2n}}e^{-n(2n)} \stackrel{n\to\infty}\longrightarrow0,$$
so that $f_n$ converges uniformly to zero.
A: If $x=0$, then $f_n(x)=0$ and therefore $\lim_{n\to\infty}f_n(x)=0$.
And if $x\in(0,1]$, then$$\lim_{n\to\infty}ne^{-nx^2}=\lim_{n\to\infty}\frac n{e^{nx^2}}=0$$and therefore $\lim_{n\to\infty}f_n(x)=0$ in this case too.
Concerning your approach, I don't see why we should have $ne^{-x}<\varepsilon$.
