Convergence of sequence of functions in metric spaces Let $f_n(x)=\frac{nx}{nx+1}$

$a)$ Show that in $C_{[0,2]}$ $$\lim_{n \to \infty}{f_n(x)}=1$$
$b)$ Does $f_n$ converge in $C_{[0,1]}?$

Here is my attempt:
$a)$ We need to show that $d(f_n,f)<\epsilon$ where $f$ is the constant function $f(x)=1$. Following previous example in my book I assume that the metric here is the supremum metric (Is this good guess if the metric is not specified?). Using the definition of the metric:
$$\sup_{x \in C_{[0,2]}}   {\left \lvert {f_n(x)-f(x)} \right \rvert} = \sup_{x \in C_{[0,2]}} {\left \lvert {\frac{nx}{nx+1}-1} \right \rvert}= \sup_{x \in C_{[0,2]}} {\frac{1}{nx+1}}=1$$
And we get a constant as a result. Not sure how to go from here. I was hoping to get something dependent on $n$ and that way we can find the appropriate term after which $d(f_n,1)<\epsilon$ for all $\epsilon>0$. 
Exactly the same can be done for $b)$ although not sure whether that proves that it converges (since we have to guess the limit first).

Note that for $x\neq 0$ we have that
$$
\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\frac{x}{x+n^{-1}}=1
$$
and clearly $f_n(0)\to 0$ as $n\to \infty$. To note that the convergence is not uniform observe that
$$
f_n(1/n)=\frac{1}{2}.
$$
