sequence of functions, limits and uniform convergence

Question

Let $f_n(x)=\frac{x}{1+nx^2}.$

(a) Find $f(x)=\lim_{n\to\infty}f_n(x)$ and $g(x)=\lim_{n\to\infty}f'_n(x)$. For which x, if any, is it true that $f'(x)=g(x)$?

(b) Prove that $|f_n(x)|\leq\frac{1}{2\sqrt{n}}$ for all $x\in R$. Do the $f_n$ converge uniformly? Why or why not?

For(a), I calculate the limit of $f_n$ and $f'_n$, and I got zero for both. Does that mean for every x, $f'(x)=g(x)$?

For(b), I have a hint to find the local extrema. But I still have no idea how to do that.

Answer 1

$\lim f_n(x)=0$ for all $x$ but $\lim f_n'(x)=0$ only for $x \neq0$. For $x=0$ the limit becomes $1$. Hence it is not true that $f'=g$.

$f_n'(x)=0$ gives $1-nx^{2}=0$ or $x=\pm \frac 1 {\sqrt n}$. The maximum value of $f_n(x)$ is attained at these points and the maximum value of $|f_n(x)|$ is $\frac 1 {2\sqrt n}$. Hence $|f_n(x)| \leq \frac 1 {2\sqrt n}$ which proves that $f_n \to 0$ uniformly.

Answer 2

(b): $|f_n(x)|\leq\frac{1}{2\sqrt{n}} \iff 1-2\sqrt{n}|x| +n|x|^2 \ge 0 \iff (1-\sqrt{n}|x|)^2 \ge 0.$