How to show failure of uniform convergence on a set Exercise Show that the sequence of functions $\langle f_n \rangle$ defined by $f_n(x)=\frac{nx}{nx+1}$ fails to converge uniformly to a function $f$ on the set $[0, \infty)$.
I understand that the tricky part here is the $0$; the $0$ is what makes it fail to be uniformly convergent. I'm having trouble showing it, however. I understand that to show uniform convergence, we must show
$$\exists \varepsilon > 0 \forall N \in \mathbb{N} \, \exists n \geq N \, \exists x \in [0,\infty): |f_n(x)-f(x)| \geq \varepsilon$$
I'm thinking the Squeeze Theorem for Limits may be the way to go here. The idea would be to use the fact that
$$ x \in [0, \infty) \Rightarrow 0 \leq x < M \, , \, M \geq x \Rightarrow f_n(0) \leq f_n(x) \leq f_n(M)$$
and eventually arrive at some contradiction or find some value $\varepsilon > 0$ that the $f_n$'s must be greater than or equal to.
Can anyone help me get started on how I should approach this problem?
 A: Do you know that if a sequence $f_n$ converges uniformly to $f$, it also converges pointwise to $f$? And that if the functions $f_n$ are continuous and converge to $f$ uniformly, then $f$ is also continuous? Due to these facts, the two first avenues of attack for showing that a sequence does not converge uniformly should usually be to:

*

*Look if it even converges pointwise (if it doesn't, it also won't do so uniformly), and

*if it does, examine the continuity of all involved functions. If the $f_n$ are continuous but their pointwise limit is not, then convergence is not uniform.

Following this advice, you should note that $f_n(0)=0$ for all $n$, so the pointwise limit at $0$ is clearly $0$. What can you say about the limit of $f_n(x)$ for fixed $x>0$? Is the resulting function $f$ continuous? Are the functions $f_n$?
A: Let $f$ be the limit function. Note that $f(0) = 0$ and $f(x) = 1$ for $x>0$.
Note that $f_n({1 \over n}) = {1 \over 2}$ for all $n$.
In particular, $|f_n({1 \over n}) - f({1 \over n})| = { 1\over 2}$ for all $n$.
