For a piecewise function $g_n(t)$, find $g(t)=\lim\limits_{n\rightarrow \infty}g_n(t)$. I think that I have found a somewhat unorthodox way of doing this question. I have written the question and my solution and finger-crossed it will be okay!

Consider the space $C=[0,1]$ of continuous functions equipped with a supremum norm $\left\|\cdot\right\|_{\infty}$. For each $n=1,2,...$ let the function $g_n \in C[0,1]$ be defined by the rule:
$$   
g_n(t) = 
     \begin{cases}
       2nt &\quad\text{if } 0\le t \le \frac{1}{2n};\\
       2 -2nt,  &\quad\text{if } \frac{1}{2n}\le t \le \frac{1}{n};\\
       0, &\quad\text{if } \frac{1}{n} \le t \le 1.\\
     \end{cases}
$$
(i) Find $g(t) =\lim\limits_{n\rightarrow \infty}g_n(t)$ for all $t\in [0,1]$. Briefly explain how you arrived at your answer.
(ii) Find $g(t) =\lim\limits_{n\rightarrow \infty}\|g_n(t)\|_{\infty}$. Briefly explain how you arrived at your answer.
(iii) Is it true that $\lim\limits_{n\rightarrow \infty}\|g_n - g\|_{\infty}=0$? Justify your answer.

My answer:
I rewrote the rule as 
$$   
f(x) = 
     \begin{cases}
       2x &\quad\text{if } 0\le x \le \frac{1}{2};\\
       2 -2x,  &\quad\text{if } \frac{1}{2}\le x \le 1;\\
       0, &\quad\text{if } 1 \le x \le n,\\
     \end{cases}
$$ where I have redefined $x=nt$.
(i) Therefore, for the first question I get $g(t)=0$ as $f(x)\rightarrow 0$.
(ii)  $g(t) =\lim\limits_{n\rightarrow \infty}\|g_n(t)\|_{\infty} =\lim\limits_{n\rightarrow \infty}\|f(x)\|_{\infty}= \lim\limits_{n\rightarrow \infty}1 = 1$
(iii)  $\lim\limits_{n\rightarrow \infty}\|g_n - g\|_{\infty}= \lim\limits_{n\rightarrow \infty}\|g_n - 0\|_{\infty}=\lim\limits_{n\rightarrow \infty}\|g_n\|_{\infty}= \lim\limits_{n\rightarrow \infty}1=1$, by the above. 

You explanation of (i) is wrong. $f(x) \not\to 0$. $f(x)$ will always consist of that same hump between $0$ and $1$ regardless of the size of $n$. The only change with $n$ is to enlarge the domain of $f$ by making the extended values $0$.
However, you have the right idea. $g(t) \to 0$ as $n\to \infty$. I suggest you fix a value of $t$ and just look at what happens at that fixed value as $n \to \infty$.
For (ii), you are correct except for the misstatement at the front. $g(t) \ne \lim\limits_{n\to \infty}\|g_n(t)\|_{\infty}$
On (iii), just make sure that you answer the question actually asked. You have the justification correct.
