In the following question I am trying to show that the sequence of derivatives $fn'$ converges uniformly on $R$
$$f_n(x)=\frac{nx^2+2x}{n}$$
So the first step would be to calculate the derivative with respect to $x$.
So then,
$$f_n'=\frac{2xn+2}{n}$$
Then taking the limit we get that,
$$lim_{n->\infty}\frac{\frac{2x}{n}+\frac{2}{n}}{\frac{n}{n}}=|2x|$$
My first question would be why is it $|2x|,$ I understand the calculation but why don't you consider when $x=0$? Is that only when you consider the pointwise limit?
Then for my workings to prove that it converges uniformly,
$$|\frac{2xn+2}{n}-2x|=\frac{2}{n}$$
Then for $\epsilon>0$ choose a $N$ such that $\frac{2}{N}<\epsilon$ and with $n>N$ then,
$$|\frac{2xn+2}{n}-2x|=\frac{2}{n}<\frac{2}{N}<\epsilon$$
Therefore the sequence of derivative converges uniformly.
So my two questions are, about the limit calculation and whether not to consider when $x=0$, and if my workings for the second part are correct.
Thanks!
The sequence of derivatives is given by
$$ f_n'(x) = 2x + \frac{2}{n}. $$
As $n$ goes to infinity, this sequence clearly converges pointwise to $2x$ (not $|2x|$). Uniform convegence is also simple because
$$ \sup_{x \in \mathbb{R}} |f_n'(x) - 2x| = \sup_{x \in \mathbb{R}} \left| \frac{2}{n} \right| = \frac{2}{n} \to 0. $$