1. another limit function question..

f_n(x)= 1 ,1<=x<n
f_n(x)= 0 ,x>n

why why whyyyy the border function is 1

i cant see the reason

2. Originally Posted by transgalactic
f_n(x)= 1 ,1<=x<n
f_n(x)= 0 ,x>n

why why whyyyy the border function is 1

i cant see the reason
You mean the limit function, as in your title. Also your definition isn't complete. Do you mean $\displaystyle f_n(x)= 1$ for $\displaystyle x\le n$? Or possibly, $\displaystyle f_n(x)= 1$ for x< 1 and $\displaystyle f_n(x)= 0$ for $\displaystyle x\ge n$? Both give the same limit but without an "=" in one of them some of the functions are not defined for some x. I'm going to assume it is $\displaystyle x\le 1$.

This is "point-wise" convergence. We look at an individual value of x, say $\displaystyle x= x_0$ and the convergence of the numerical sequence {f_n(x_0)}. For example, if x= 2, then $\displaystyle f_1(2)= 0$ because 2> 1. $\displaystyle f_2(2)= 1$ because 2= 2. But $\displaystyle f_3(2)= 0$ because 2< 3 and, for all n> 3, 2< n so $\displaystyle f_m(2)= 0$. The sequence $\displaystyle \{f_n(2)\}$ is 1, 1, 0, 0, 0, 0, ... which converges to 0.

In fact, for any x, there exist some positive integer, N, such that x< N (that's the "Archimedian property" of the positive integers). For any such N, $\displaystyle f_N(x)= 0$ and for all n> N, $\displaystyle f_n(x)= 0$. The sequence {f_n(x)} consists of some finite number of "1"s followed by an infinite sequence of "0"s and that converges to 0.