for which $x$, is $f'(x)=lim_{n \to \infty}f_n'(x)$?

Question

let $f_n(x)=\dfrac{x}{1+nx^4} \quad n=1,2... \quad$ and $f(x)=\lim_{n \to \infty}f_n(x)$
I have to determine for for which $x$, is $f'(x)=\lim_{n \to \infty}f_n'(x)$?
Here is what I have done:
$f_n'(x)=\dfrac{1-3nx^4}{(1+nx^4)^2}$ and I have found that $f(x)=\lim_{n \to \infty}\dfrac{x}{1+nx^4}=0.$
Now I don't really understand the question

"determine for which $x$, is $f'(x)=\lim_{n \to \infty}f_n'(x)$?"
Any suggestions?

Answer 1

Assuming you calculated $f_n'$ correctly.

As you pointed out, $f(x)=0$, so $f'(x)=0$. Now $\lim\limits_{n \to \infty} f_n'(x) = \lim\limits_{n \to \infty} \frac{1-3nx^4}{(1+nx^4)^2}$; this is $0$ if $x\neq0$ and $1$ if $x=0$.

Therefore the answer is for every $x\neq0$.