All I need a confirmation if my procedure is correct.
$1. \rightarrow$ $$f_n(x)= n\log\bigg(\frac{1+nx}{nx}\bigg),$$ $x \in (0, \infty).$
$2. \rightarrow$ $$f_n(x)= n(x^\frac{1}{n}-1),$$ $x\in [1,a], a>1.$
From $1.$, I have
$$|f'_n(x)-g(x)|<\epsilon$$
$$\Rightarrow n> \frac{1}{x}(\frac{1}{x\epsilon}-1),$$ $x \in (0, \infty).$
At $x=0$ the sequence has a point of discontinuity and as $x \rightarrow \infty$, $f_n$ approaches to $0$ but in $(0,\infty)$ the $f_n(x)$ is continuous and $0<n\log(1+\frac{1}{nx})<(1+\frac{1}{nx})^n<1$ for large $n$ and an $M_n$ test gives $$\max_{x\in (0, \infty)}|f'_n(x)-g(x)|=0,$$ where $g(x)=\lim_{n \rightarrow \infty}f'_n(x).$
By applying the continuity and boundedness can I say $f_n(x)$ is uniformly convergent?
From $2.$, clearly I have that $f_n(x)$ is bounded and decreasing in $(-\infty, 0]$ and $f_n(x)$ has discontinuties as $n \rightarrow \infty$ since $$\lim_{n \rightarrow \infty} n(a^\frac{1}{n}-1)=0,$$ $a>1.$
So $f_n(x)$ is continuous and $0\leq f_n(x)<n$ but a little algebra shows that none of $f_n(x)$ or $f'_n(x)$ approaches to 0, rather they have discontinuity for larger $n$. Is is enough to say that $f_n(x)$ is non-uniform?
Any help is appreciated.