Sequence of real numbers: #
Consider sequence of real numbers \begin{align} & \{y_{n}\}=\left\{ 1,\frac{1}{2},\frac{1}{3},…,\frac{1}{n},…\right\} \nonumber\\ & \end{align}
Then clearly $$\{y_{n}\}\to 0 ~as ~n\to \infty~or~\lim\limits_{n\to \infty}y_{n}=0 $$
And, by the definition of limit of a sequence, for any positive number \(\epsilon~,~\exists \) a positive integer \(k \) such that \begin{align} & |y_{n}-0|<\epsilon~\forall~n\ge k \nonumber\\ \implies & |y_{n}|<\epsilon~\forall~n\ge k \nonumber\\ & \end{align}
Sequence of real valued functions: #
Now let us a sequnce of real valued functions \begin{align} &\{f_{n}(x)\}=\left\{ x,\frac{x}{2},\frac{x}{3},…,\frac{x}{n},…\right\} \nonumber\\ & \end{align} defined on the interval \( [-1,1] \).
More explicitly, \begin{align*} & f_{1}(x)=x \\ & f_{2}(x)=\frac{x}{2}\\ & f_{3}(x)=\frac{x}{3}\\ & ……………..\\ & f_{n}(x)=\frac{x}{n}\\ & …………….. \end{align*}
Now if we put \(x=1\), in the sequence of functions (2), then we get the sequnce of real numbers \( \{ f_{n}(1)\} \) which is exactly equal to the sequence \( \{y_{n}\} \) as shown in (1). So, we have \begin{align} & \{f_{n}(1)\}\to 0 ~as ~n\to \infty \nonumber\\ ~or~&\lim\limits_{n\to \infty}f_{n}(1)=0 \nonumber\\ & \end{align}
Then for positive number \(\epsilon>0~\exists \) a positive integer \(k \) such that \begin{align} & |f_{n}(1)-0|<\epsilon~\forall~n\ge k \nonumber\\ \implies & |f_{n}(1)|<\epsilon~\forall~n\ge k \nonumber\\ & \end{align}
