Prove that the sequence of functions $g_{n}\in C[0,4]$ I want to show that the function defined by $g_n:[0,4]\to \Bbb{R}$, defined by\begin{align} g_n(t)=\begin{cases}0,& \text{if}\;0\leq t\leq 2,\\\dfrac{n}{2}(t-2),& \text{if}\;2\leq t\leq 2+\frac{2}{n},\\1 &\text{if}\;2+\frac{2}{n}\leq t\leq 4.\end{cases} \end{align}
is continuous and Cauchy.
MY TRIAL
The sequence of functions is clearly continuous by Pasting Lemma. Next, we show that the sequence of functions is Cauchy. 
So, let $m,n\in\Bbb{N},$ be given such that $m\geq n.$ Then,
$$|g_m(t)-g_n(t)|=|0-0|+\left|\dfrac{m}{2}(t-2)-\dfrac{n}{2}(t-2)\right|\left|1-1\right|=0<\epsilon$$
I don't know if this is right. If not, kindly help please!
 A: What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”? 
This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 \leq x \leq 2$ and $h(x)=1$ otherwise. 
Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.
Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $\mathbb{R}$ is complete for the max-norm.
$g_n\left(2+\frac{2}{2n}\right)=\frac{n}{2}\frac{2}{2n}=\frac{1}{2}$, while $g_{2n}\left(2+\frac{2}{2n}\right)=1$. 
Thus the max-norm of $g_n-g_{2n}$ is not lower than $\frac{1}{2}$, hence the sequence is not Cauchy.
A: A sequence is Cauchy if for every $\epsilon$ there exists $N \in \mathbb{N}$ such that $\forall n, m > N$ we have $|g_n-g_m|< \epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
\begin{align}
(g_m-g_n)(t)= \begin{cases}
0 & if\  0 \leq t \leq 2
\\(\frac{m}{2}-\frac{n}{2})(t-2) & if \ 2 < t \leq 2+\frac{2}{m} 
\\1-\frac{n}{2}(t-2) & if\ 2+\frac{2}{m} < t \leq 2+\frac{2}{n} 
\\ 0 & if\ 2+\frac{2}{n} < t \leq 4
\end{cases}
\end{align}
Now $||g_m-g_n||_1=\int_2^{2+\frac{2}{m}}|(\frac{m}{2}-\frac{n}{2})(t-2)|dt+\int_{2+\frac{2}{m}}^{2+\frac{2}{n}}|1-\frac{n}{2}(t-2)|dt$ 
If we compute explicitly the integrals we get $||g_m-g_n||_1=\frac{m-n}{m^2}+\frac{(m-n)^2}{nm^2}=\frac{m-n}{mn}<\frac{m}{mn}=\frac{1}{n}$
Let $N \in \mathbb{N}$ such that $N > \frac{1}{\epsilon}$ and we have $||g_m-g_m||_1<\frac{1}{n}<\frac{1}{N}<\epsilon$ and this proves the sequence is Cauchy
