For $n$ a positive integer, let $f_n(x)$ be the continuous function $\frac{1}{1+nx}$ with domain the positive real numbers. Let $f (x)$ be the pointwise limit of the sequence $\{ f_n(x)\}_{n\geq1}$ of functions. On which of the following intervals is the convergence $f_n\to f$ uniform? Choose all the correct options:
(a) $(0,1)$
(b) $(1,2)$
(c) $(2,1)$
(d) None of the above.
My attempt:
We have $f_n(x)=\frac{1}{1+nx}.$ Therefore the pointwise limit is $f(x) = 1$.
Now using $\epsilon-\delta$ definition, we have to find a $N$ which only depends on $\epsilon$ but not $x.$
$|\frac{1}{1+nx}-1| <\epsilon$ $\implies \frac{nx}{1+nx}<\epsilon \implies n > \frac{-\epsilon}{x(1-\epsilon)}.$
I am stuck here, please help.
Actually, for each $x>0$, $\lim_{n\to\infty}\frac1{1+nx}=0$. The convergence is uniform on $(1,2)$, because if $x\in(1,2)$, then$$\left\lvert\frac1{1+nx}-0\right\rvert=\frac1{1+nx}<\frac1{1+n},$$since $x>1$. So, given $\varepsilon>0$, take $N\in\mathbb N$ such that $\frac1{1+N}<\varepsilon$ and then$$(\forall n\in\mathbb N)(\bigl(\forall x\in[0,1]\bigr):\left\lvert\frac1{1+nx}-0\right\rvert<\frac1{1+n}\leqslant\frac1{1+N}<\varepsilon.$$
And the convergence is not uniform on $(0,1)$, $(\forall n\in\mathbb N):f_n\left(\frac1n\right)=\frac12$.
Logically, $(2,1)=\{x\in\mathbb R\mid2<x<1\}=\emptyset$. So, the convergence is uniform there.