Definition: #
Let \(\bf \{ f_{n}(x)\} \) be a sequence of real velued functions, defined on \(\bf E\subseteq \real \). The sequence \(\bf \{ f_{n}(x)\} \) is said to be \(\fcolorbox{red}{white}{\bf pointwise convergent} \)to a real valued function \(\bf f(x) \), defined on \(\bf E\subseteq \real \), if for any positive number \(\bf \epsilon~,~\exists \) a positive integer \(\bf k \) such that \begin{align} & \bf|f_{n}(x)-f(x)|<\epsilon~\forall~n\ge k~and~x\in E \nonumber \\ & \end{align} And is denoted by $$\bf \lim\limits_{n\to \infty}f_{n}(x)=f(x) $$
Example-1: \(\bf f_{n}(x)=x^{n} \) on \(\bf x\in [0,1]\) #
Solution:
We know that the sequence of real number \(\bf \{y_{n} \} \), where \(\bf y_{n}=x^{n}~\forall~|x|\text{\textless} 1 \), converges to \(\bf 0 \). Then $$\bf \lim\limits_{n\to \infty}y_{n}=0 $$ Thus we have $$ \bf \lim\limits_{n\to \infty}f_{n}(x)=0~when~0\le x \text{\textless} 1$$ Also, for \(\bf x=1\), we have the sequence \(\bf \{ f_{n}(1) \}=\{1,1,….\} \). Therefore we have $$ \bf \lim\limits_{n\to \infty}f_{n}(x)=1~when~x=1$$ Hence $$\bf \lim\limits_{n\to \infty}f_{n}(x)=f(x)~when~x\in [0,1]$$ where $$\bf f(x)=\begin{cases} \bf 0 &\text{\bf if } \bf x\in [0,1) \\ \bf 1 &\text{\bf if } \bf x=1 \end{cases} $$
