I have solved the following exercise and I would like to know if I have made any mistakes:
Let $g_n(x)=\frac{nx+x^2}{2n}$ and set $g(x)=\lim g_n(x)$. Show that $g$ is differentiable in two ways:
(a) Compute $g(x)$ by algebraically taking the limit as $n\to\infty$ and then find $g'(x)$.
(b) Compute $g'_n(x)$ for each $n\in\mathbb{N}$ and show that the sequence of derivatives $(g'_n)$ converges uniformly on the interval $[-M,M]$. Conclude $g'(x)=\lim g'_n(x)$.
(c) Repeat parts (a) and (b) for the sequence $f_n(x)=\frac{nx^2+1}{2n+x}$.
My solution:
(a) For fixed $x\in\mathbb{R}: \frac{nx+x^2}{2n}=\frac{nx}{2n}+\frac{x^2}{2n}=\frac{x}{2}+\frac{x^2}{2n}\xrightarrow[]{n\to\infty}\frac{x}{2}=:g(x)$ so $g'(x)=\frac{1}{2}$;
(b) $g'_n(x)=\frac{1}{2}+\frac{x}{n}$ and $|g'_n(x)-\frac{1}{2}|=\frac{|x|}{n}\leq\frac{M}{n}<\varepsilon$ for $n>\frac{M}{\varepsilon}$ so $g'_n$ is uniformly convergent to $\frac{1}{2}$ on each interval $[-M,M]$ and since $0\in [-M,M]$ for every $M\in\mathbb{R}$ and $g_n(0)=0\xrightarrow[]{n\to\infty}0$, so we can conclude that $g_n$ converges uniformly on $[-M,M]$ and $g=\lim_{n\to\infty} g_n$ is differentiable and satisfies $g'=\lim g'_n$.
(c) For fixed $x\in\mathbb{R}: \frac{nx^2+1}{2n+x}=\frac{1}{2}\frac{x^2 +\frac{1}{n}}{1+\frac{x}{2n}}\xrightarrow[]{n\to\infty}\frac{x^2}{2}=:f(x)$ and $f'(x)=x$;
$f'_n(x)=\frac{4n^2x+nx^2-1}{(2n+x)^2}$ and $|f'_n(x)-x|=|\frac{x^3+3nx^2+1}{(2n+x)^2}|\leq\frac{x^2 |x+3n|+1}{4n^2}\leq\frac{M^2 (M+3n)+1}{4n^2}\xrightarrow[]{n\to\infty}0$ so $f'_n$ converges uniformly to $x$ on $[-M,M]$ and since $0\in [-M,M]$ and $f_n(0)=\frac{1}{2n}\xrightarrow[]{n\to\infty}0$ so we can conclude that $f_n$ converges uniformly on $[-M,M]$ and $f=\lim f_n$ is differentiable and satisfies $f'=\lim f'_n$.