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So I need to show that $f_n(x) = \frac{nx}{1+nx}$ is not uniformly convergence on $[0,\infty)$. I found a way of showing this, but I'm not sure whether this is correct. I'd much appreciate it if someone could either confirm whether I can use this or show the error I made!

$\exists \epsilon > 0: \forall N\in\mathbb{N}: \exists n\geq N : \exists x \in [0,\infty): |\,f_n(x) - f(x)| \geq \epsilon$

Choose for $\epsilon = \frac{1}{2}$, choose for $x = 0$ and choose for $n = N$.

$ | \frac{nx}{1+nx} - 1| = | \frac{nx-1-nx}{1+nx}| = |\frac{1}{1+nx}| = \frac{1}{1} = 1 \geq \frac{1}{2} = \epsilon $

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    The pointwise limit is $1$. It is enough to show that $\sup |f_n - 1|$ does or does not go to zero if you want to test uniform convergence. – 2017-01-17
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    An easier approach than disproving uniform continuity would be to state that the pointwise limit is not continuous, as $f_n(x)\rightarrow 0$ for $x=0$, but $f_n(x)\rightarrow 1$ for any other $x\in(0,\infty)$. Because each $f_n(x)$ is continuous, uniform continuity would guarantee that the limit function is continuous as well. – 2017-01-17

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You are on the right track. The pointwise limit at $x=0$ is $f_n(0)\rightarrow 0.$ So the sequence converges pointwise to $1,$ except at $x=0,$ where it converges to zero. So you can't choose $x=0$... the functions are always zero there, so the distance $|f_n(0)-f(0)|$ is always zero.

So limit our consideration to $x>0.$ As you computed, $$|f_n(x)-f(x)| = \frac{1}{1+nx}.$$

Let $N>1$ be chosen and just pick any $n>N$. Then we can always pick an $x>0$ such that $\frac{1}{1+nx} > 1/2$ (just take $x<\frac{1}{n}).$