Show that $f(x)=n/(1+nx)$ does not uniformly converge on $(0, \infty)$

Question

I have proved that this sequence of functions is uniform convergent on $[1,\infty)$ by choosing $N>1/\epsilon$ (correct me if i am wrong).

Now need to show that it is not uniform convergent on $(0,\infty)$. What should be the choice for $N$?

Answer 1

Hint: If $x=\frac1{n^2}$, then$$\frac n{1+nx}=\frac n{1+1/n}=\frac{n^2}{n+1}.$$

Answer 2

Pointwise, it converges to $x \mapsto \frac{1}{x}$. If it was uniformly convergent on $(0,\infty)$, we would have $$ \mathop {\lim }\limits_{n \to + \infty } \mathop {\sup }\limits_{x > 0} \left| {\frac{1}{x} - \frac{n}{{1 + nx}}} \right| =0. $$ But $$ \mathop {\sup }\limits_{x > 0} \left| {\frac{1}{x} - \frac{n}{{1 + nx}}} \right|\mathop \ge \limits^{x = 1/n} \left| {n - \frac{n}{2}} \right| = \frac{n}{2}. $$ A usual way showing that $f_n(x)\to f(x)$ is not uniform on $A$ is to give a lower bound on $\sup_{x\in A} |f(x)-f_n(x)|$ which does not tend to $0$ when $n\to +\infty$. This is generally done by choosing an appropriate $x$ in $A$ as a function of $n$ (in our example $x=\frac{1}{n} \in (0,\infty)$).