$f_n(x) = \dfrac{nx^2 + 1}{2n + x}$, assume $f_n$ converges to $f$ and $f'_n$ converges to $g$ pointwisely. Then $f' = g$ on a compact subset of reals

Question

I showed that $f_n$ converges to some $f$ uniformly. Now I need to show that $f'_n$ converges to $g$ uniformly such that $g=f′$, but I do not know how to show it. Because I want to utilize the following theorem: Let $f_n:[,] \to ℝ$ be a sequence of differentiable functions whose derivatives $f'_n$ are continuous. If $f_n$ converges uniformly to $f$ and $f'_n$ converges uniformly to $g$, then the limit $f$ is differentiable, and its derivative is $′=$

Answer 1

As it was pointed out, you can show this directly. You already showed that $f_n(x) \to \frac 12 x^2$, uniformly on $[a,b]$. So, you just need to verify that $$ \sup_{x\in[a,b]}|f'_n(x)-x| \to 0 \quad(n \to \infty), $$ which is true, since \begin{align*} |f'_n(x)-x| = \frac{|3 n x^2+x^3+1|}{(2 n+x)^2} = \frac{n(3x^2+x^3/n+1/n)}{n^2(x^2/n^2+4x/n+4)} \to 0 \quad (n\to \infty, x \in [a,b]). \end{align*}