Uniform convergence of $\frac{x^n}{1+x}$ and $\frac{x^n}{1+nx}$

Question

if $$f_n(x)=\frac{x^n}{1+x}$$and $$g_n(x)=\frac{x^n}{1+nx}$$ ,then on the interval $[0,1]$,check the uniform convergence of above functions ,if:- $$(1.)x\in \mathbb{N}$$ $$(2.)x\in \mathbb{R}$$ My approach:-
$(1.)$ when $x \in \mathbb{N}$:-
$$\lim_{n\to \infty}f_n(x)=\infty$$,because for both $0 $ & $1$ ,$f_n(x) $ gives different values;....so $f_n(x)$ is not uniformly convergent Now for $g_n(x)$:-
for $$x=0, f_n(x)=0$$and for$$x=1,f_n(x)=\frac{1}{1+n} $$such that $\lim_{n\to\infty} f_n(x)=0$ for both $x=0$ and $x=1$ Now
As $$\lim_{n \to \infty} sup |g_n(x)-0|=0$$ thus $g_n(x)$ is uniformly convergent but $f_n(x)$ is not uniformly convergent for $x \in \mathbb{N}$

$(2.)$ when $x \in \mathbb{R}$ ,i think that both sequences diverge for this set
Is my approach correct or not?

Answer 1

For $$(2.) x\in \mathbb{R}$$
both $f_n(x)$ and $g_n(x)$ are converging to 0 as follows:- since $0 \le x \le 1$ $$h(x)=\lim_{n\to\infty}f_n(x)=\lim_{n \to \infty}g_n(x)=0$$ Now, we have to find $\lim_{n\to\infty}sup|f_n(x)-h(x)|$ and $\lim_{n\to\infty}sup|g_n(x)-h(x)|$ as follows:-
$x$ ,such that $\frac{d}{dx}|f_n(x)-h(x)|=0$ i.e $x=1$
and at $x=1 ,f_n(x)=\frac{1}{2}$
Similarly $x$ ,such that $\frac{d}{dx}|g_n(x)-h(x)|=0$ i.e. $x=0$
and at $x=0 ,g_n(x)=0$
so$$\lim_{n\to\infty}f_n(x)_{x=1}=\frac{1}{2}\ne0$$and $$\lim_{n\to\infty}g_n(x)_{x=0}=0$$ so $f_n(x)$ does not converge uniformly but $g_n(x)$ converges uniformly.