Show that the sequence of functions $\langle f_n \rangle$ defined by $$f_n(x) = \frac{nx}{nx+1} \, ; \, n \in \mathbb{N}$$ fails to converge uniformly on $[0, \infty) .$
The name of the game here is to observe that the converge is only non-uniform for $x=0$, i.e. the convergence is uniform on $[a, \infty)$ for $a> 0$. I already showed this in the first part of my proof. but my proof for $x \in [0, \infty)$ so far looks like this:
Proof
Consider the same inequality as in the previous problem. But now let $x \in [0, \infty)$.
Recall again that for, $x \in [a, \infty)$, $a>0$, we have that $\displaystyle\lim_{n \to \infty} f_n(x) = f(x) = 1$.
Since we have shown uniform convergence at every $x \in [a, \infty)$ for every $f_n$, we need only show the case when $x \in [0, \infty)$, namely when $x = 0$.
Let $t \in (0, x)$, i.e. $0 < t < x$. Since $t < x$, we have $\frac{1}{x} < \frac{1}{t}$ and similarly, $\frac{1}{nx+1} < \frac{1}{nt+1}$
I would like to essentially get to a point where I can take $\displaystyle\lim_{t \to 0^+}$ and show that $|f_n - f(x)| \geq \varepsilon$ but I'm having trouble because
$$\frac{nx}{nx+1} \not< \frac{nt}{nt+1}$$
Any advice or pointers?
Or would it be a better idea to just plug $0$ into any $f_n$ and somehow develop $\varepsilon$ based on the fact that $f_n(0) = 0$?
The sequence $(f_n)_{n\in\Bbb N}$ converges pointwise to $1$. Therefore, if it converged uniformly on $[0,\infty)$, it would converge uniformly to $1$. But$$(\forall n\in\Bbb N):f_n\left(\frac1n\right)=\frac12.$$