Showing that $g_n(x)=x\cdot \frac{1}{nx+1}$ converges uniformly on $(0,1)$

Question

I want to show that $g_n(x)=x\cdot \frac{1}{nx+1}$ converges uniformly on $(0,1)$, where $n \in \mathbb{N}$.

Let $f_n(x)=\frac{1}{nx+1}$. Note that $f_n'(x)=\frac{-n}{(nx+1)^2}$, so $f_n$ is always decreasing over $(0,1)$. My idea is to show that $f_n$ converges uniformly on $[\epsilon,1)$ for all $\epsilon\in (0,1)$. Then I can use the fact that $g(x)=x$ is a continuous function and that given $\epsilon >0$, I can take $\delta=\epsilon$ so that $|g(x)-g(y)|=|x-y|<\epsilon$.

I can split up $(0,1)=(0,\epsilon)\cup [\epsilon,1)$. I know how to show that $g_n(x)=g(x)f_n(x)=x\cdot \frac{1}{nx+1}$ converges uniformly on $[\epsilon,1)$, but then I get stuck with showing uniform convergence on $(0, \epsilon)$. I am not very good at this and I feel like I am going in circles. Please help me! Thank you!

Answer 1

$0<\frac{x}{nx+1}<\frac{x}{nx}<\frac{1}{n}$ for all $x\in (0,1)$.

So if you pick $\epsilon>0$ you just need to pick an $N$ so that $N>\frac{1}{\epsilon}$.

Answer 2

For all $n \in \mathbb{N}$, $$\sup_{(0,1)}\{g_n(x):x\in(0,1)\}= \Big|\frac{x}{nx+1}\Big|_{(0,1)} =\frac{1}{n+1}.$$ Hence $$\lim_{n \to \infty} \Big|\frac{x}{nx+1}\Big|_{(0,1)}=\lim_{n \to \infty} \frac{1}{n+1}=0.$$