Find the limit function $f$ for $\{F_n\}$ where $F_n=nxe^{-nx^2}$
My solution:
$$\lim_{n\to\infty}nxe^{-nx^2}=x\lim_{n\to\infty}nxe^{-nx^2}=x\lim_{n\to\infty}\dfrac{1}{x^2e^{nx^2}}=\dfrac{1}{x}\lim_{n\to\infty}e^{-nx^2}=0, x\neq0$$
$$F_n=0\text{, for }x=0$$
Hence $\{F_n\}$ converges point wise to zero on $\mathbb{R}$
The result is right, but I absolutely don't understand your computations. What did you do at the first and second equation and why?
Either you have a theorem which claims that the exponential function dominates polynomials or you can see it this way:
For $y>0$ consider $$ e^y=\sum_{k=0}^\infty\frac1{k!}y^k\geq\frac12y^2. $$ ($\frac12x^2$ is the third term of the sum. Drops all other term, which are all positive since $y>0$)
Because of $nx^2>0$ for $x\neq 0$, this yields \begin{align} 0\leq \lim_{n\to\infty} \left|nxe^{-nx^2}\right|=\lim_{n\to\infty}\frac{n|x|}{e^{nx^2}}\leq \lim_{n\to\infty}\frac{n|x|}{\frac12(nx^2)^2} =\lim_{n\to\infty}\frac{2}{n|x|^3}=0. \end{align} We get $\lim_{n\to\infty}nxe^{-nx^2}=0$ for $x\neq 0$ as stated.
(The first inequality is obvious since the absolut value is always nonnegative. Because of $-nx^2<0$, I change at the first equation the product to a fraction and used that the exponential term and $n$ are positive. Next I used the inequality from above. Finally I simplified the term to see that the limit is $0$. Since the left and right end are $0$, all inequalities become equalities and we get our statement.)