Show that $$f_n(x) = \frac{nx}{nx+1}$$

converges uniformly to a function $f$ on domain $[a, \infty)$ where $a>0$.

Solution Verification Requested

Suppose $f_n: [a, \infty) \to \mathbb{R}$ is given by $f_n(x) = \frac{nx}{nx+1}$ .First notice that

$$f(x) = \lim_{n \to \infty}f_n(x) = \lim_{n \to \infty} \frac{nx}{nx+1} =1$$

Second, note that

$$|f_n(x) - f(x)| = \Big| \frac{nx}{nx+1} - 1 \Big| = \Big| \frac{nx}{nx+1} - \frac{nx+1}{nx+1} \Big| = \Big| \frac{-1}{nx+1} \Big| = \frac{1}{nx+1}$$

Now, let $\varepsilon > 0$. By the Archmidean Property, we can find $N \in \mathbb{N}$ so that

$$\frac{1}{Nx+1} < \varepsilon \hspace{1cm} x \in [a, \infty)$$

Suppose $n \geq N$. Since $a \leq x$, then $\frac{1}{x} \leq \frac{1}{a}$. Then we have

$$\frac{1}{nx+1} \leq \frac{1}{na+1}$$

Altogether, we have found

$$\frac{1}{nx+1} = |f_n(x)-f(x)| < \varepsilon$$

whenever $n \geq N$ and $x \in [a, \infty)$. So $f_n \to f$ uniformly.

As a final check, we check the endpoint $a$ and see that

$$|f_n(a+\frac{1}{n})-f(a+\frac{1}{n})| = \frac{1}{n(a+\frac{1}{n})+1} = \frac{1}{na+2} < \varepsilon \hspace{0.3cm} \mathrm{whenever} \hspace{0.3cm} n \geq N$$

  • $\begingroup$ @ThomasAndrews My bad, I meant to add that in at the beginning but forgot. It's fixed now $\endgroup$ Commented Jun 23, 2022 at 19:13

2 Answers 2


I think that's perfect. Take a look to this form, maybe you'll find it easier, it uses the supremum criteiron $$\sup_{x \in [a,\infty)} \lvert f_n(x)-f(x) \rvert=\sup_{x \in [a,\infty)} \left\lvert \frac{1}{1+nx} \right\rvert=\frac{1}{1+n a} $$ and the latter tends to $0$ as $n \to \infty$.


Your appeal to the Archimedian property is dubious, because you are want it true for all $x.$ The reason you can do it is that $\frac{1}{nx+1}$ has a maximal value at $x=a.$ So just pick $N$ so that: $$\frac{1}{Na+1}<\epsilon.$$

I'd simplify it further, and pick $N>\frac{1}{a\epsilon}.$ Then for $n\geq N, x\geq a,$ $nx\geq Na>0,$ and thus $$\frac{1}{nx+1}<\frac{1}{nx}\leq\frac{1}{Na}<\epsilon.$$


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