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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!

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    Why are you trying to do it in a complicated way? I think it's pretty obvious that it is uniformly convergent to $0$ from definition only, or there are some other easy ways to see this if you know a theorem or two about uniform convergence – 2017-01-21
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    It is obvious that $gn$ converges to 0 (zero function). Later just show that $Mn$=sup($gn$-0) converges to zero. It is a very simple and direct exercise. – 2017-01-21

2 Answers 2

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

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

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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 $$\lim_{n \to \infty} \Big|\frac{x}{nx+1}\Big|_{(0,1)}=\lim_{n \to \infty} \frac{1}{n+1}=0.$$