Let $f_n(x) = (x-\frac{1}{n})^2$ for $x\in[0,1]$. Does the sequence $\{f_n\}$ converge pointwise in the set $[0,1]$?
And the question asks to give the limit function as well.
So far I have:
$$f(x)=\lim_{n\rightarrow\infty}f_n(x)$$
$$f(x)=\lim_{n\rightarrow\infty}(x^2-\frac{2x}{n}+\frac{1}{n^2})$$
$$f(x)=x^2 \;\forall x\in[0,1]$$
I am stuck from here

Your approach is correct and indeed $f_{n}\to x^{2}$ pointwise on $[0,1]$.
The definition of pointwise convergence of a sequence of functions $(f_{n}(x))_{n\in \mathbb{N}}$ on a set $I\subseteq \mathbb{R}$ says, $f_{n}\to f$ pointwise on $I$ if, and only if,
$$(\color{red}{\forall x\in I})(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$
Let the sequence of functions $f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$ and let's to show that $f_{n}\to f$ pointwise on $[0,1]$, with $f(x)=x^{2}$.
Let $x\in [0,1]$ and let $\varepsilon >0$. By the Archimedean property
there exists a natural number $N>\frac{\varepsilon}{3}$. If $n>N$ we have
\begin{align*}
|f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|\\&=\left|x^{2}-\frac{2x}{n}+\frac{1}{n^{2}}-x^{2}\right|\\ &\leqslant \frac{|2x|}{n}+\frac{1}{n^{2}},\quad 0\leqslant  x\leqslant 1,\\&\leqslant \frac{2}{n}+\frac{1}{n^{2}}\\&\leqslant \frac{3}{n},\\&<3\left(\frac{\varepsilon}{3}\right),\\&=\varepsilon.
\end{align*}
Therefore $f_{n}\to f$ pointwise on $[0,1]$.
Moreover $f_{n}\rightrightarrows f$ uniformly on $[0,1]$.
The definition of uniform convegence of a sequence of functions $(f_{n}(x))_{n\in \mathbb{N}}$ on a set $I\subseteq \mathbb{R}$ says, $f_{n}\rightrightarrows f$ uniformly  on $I$ if, and only if,
$$(\forall \varepsilon >0)(\color{red}{\exists N\in \mathbb{N}})(\forall n\in \mathbb{N})(\color{red}{\forall x\in I})(n>N\implies |f_{n}(x)-f(x)|<\varepsilon.$$
Let the sequence of functions $f_{n}(x)=\left(x-\frac{1}{n}\right)^{2}$ and let's to show that $f_{n}\rightrightarrows f$ uniformly on $[0,1]$, with $f(x)=x^{2}$.
Let $\varepsilon >0$, thus choose $N\in\mathbb{N}$ such that $\frac{1}{N^{2}}<\frac{\varepsilon}{2}$ and $\frac{2x}{N}<\frac{\varepsilon}{2}$. Thus for all $n>N$ we have
\begin{align*}
|f_{n}(x)-f(x)|&=\left|\left(x-\frac{1}{n}\right)^{2}-x^{2}\right|,\\&=\left|-\frac{2x}{n}+\frac{1}{n^{2}}\right|,\\&\leqslant \frac{2x}{n}+\frac{1}{n^{2}},\\&<\frac{2x}{N}+\frac{1}{N^{2}},\\&=\frac{\varepsilon}{2}+\frac{\varepsilon}{2},\\&=\varepsilon.
\end{align*}
Therefore $f_{n}\rightrightarrows f$ uniformly on $[0,1]$.
Alternatively notice that
$$\sup_{x\in [0,1]}\left|f_{n}(x)-f(x)\right|\to 0,\quad n\to +\infty$$
Therefore $f_{n}\rightrightarrows  f$ uniformly on $[0,1]$. Then also $(f_{n}(x))_{n\in \mathbb{N}}$ is uniform Cauchy.
