Uniform and pointwise convergence of sequence in standard topology. Let $(f_{n})$ be a sequence of functions $ f_{n} \colon [-1,0) \to \mathbb{R} $ so that $ f_{n} (x) =0$, for $ x \in [-1, - \frac{1} {n}) $ and $ f_{n} (x) =nx+1$, for $ x \in [ - \frac{1} {n}, 0)$, for each $ n \in \mathbb{N} $. 
Does it converge pointwise? Does it converge uniformly? 
I think it converges pointwise to $f(x) =0$ by looking at the limit of $(f_{n})$ but I am not sure about uniform convergence. 
 A: It does converge to $0$ pointwise. If the convergence is uniform then there exits $n_0$ such that $|f_n(x)| <1/2$ for all $x $ for all $n >n_0$. In particular $|f_n(-\frac 1 {n^{2}}) | <1/2$. But this says $1-\frac 1 n <1/2$ for all $n >n_0$. We get a contradiction by letting $n \to \infty$. 
A: It converges pointwise to the 0 function. To see this, note that for any $x \in [-1, 0)$, you can find $n \in \mathbb{N}$ such that $x < -\frac{1}{n}$.
It does not converge uniformly. To see this, let for example $\epsilon =1$, and assume we can find $N \in \mathbb{N}$ such that $|f_n(x) - 0|=|f_n(x)|<1$ for all $n\ge N$ and for all $x \in [-1, -0)$. Why is this wrong?
Edit: Pointed out in comments, for uniform convergence, it needs to hold for all $x \in [-1,0)$.
A: Pointwise $f_n (x)$ converges pointwise to $f(x)=0$ in $[-1,0)$.(See other answers).
Option : 
$||f||:=\sup${ $|f(x)| x \in [-1,0)$}.
$f_n$ converges uniformly to $f$  in $[-1,0)$
$ \iff $
$\lim_{n\rightarrow \infty}||f_n-f||=0.$
With $f(x)=0$ we have
$||f_n||=\sup$ { $|f_n(x)|x \in [-1,0)$ } $=1$
Hence $\lim_{n \rightarrow \infty}||f_n|| =1$, does not converge uniformly.
