I have always tried to find these evaluations of limits of functions very very difficult. First of all, I need to know the basic steps of calculating these, is intuition a key factor in calculating these, for example
$$f_n(x)=\begin{cases}n^2x,~\text{if}~ 0\leq x\leq 1/n\\ -n^2(x-2/n), ~\text{if}~ 1/n\leq x\leq 2/n\\ 0 ~\text{if} ~2/n\leq x\leq 1 \end{cases}$$ The first step is to calculate the pointwise limit $f(x)$, but I am stuck at this step itself. I know the limit $f(x)$ means $|f_n(x)-f(x)|<\epsilon$ $\forall n\geq N_0(\epsilon,x), x\in [0,1]$, But how do I find the limit using this definition. At the moment I am not really able to grasp this concept of convergence of sequence of functions let alone finding their limit.
Finding the pointwise limit is very similar to finding the limit of a sequence, although you have to find the limit for all possible values of x. Usually it is useful to consider more cases.
In the above example we might look at two cases, case 1: $x=0$, and case 2: $x\in (0,1]$.
Case 1: By definition of $f_n$ we have that $f_n(0)=n^2 \cdot 0 = 0$ for all $n\in \mathbb{N}$, therefore $\lim_{n\rightarrow \infty} f_n(0)=0$.
Case 2: Let $x\in (0,1]$. Since $\lim_{n\rightarrow \infty} \frac{2}{n}=0$, there must exist $N\in\mathbb{N}$ such that $2/n\leq x$ for all $n\geq N$. But this means, that for $n\geq N$ we must have that $x\in(2/n,1]$, hence $f_n(x)=0$, and we can conclude that $\lim_{n\rightarrow \infty} f_n(x)=0$.
A useful trick can also be to sketch the graph of the first few functions and maybe guess at what the limit should be.