So my attempt to solve this is that i know how the behavior of the function, i.e, it will go for
$n=1: [0,1)$ and $F(x)=0$, $\forall$ $x\in[0,1)$
Then the graphic will go on the x from 0 to 1 and in 1 it will be empty but in
$n=2: [1,2)$ and $F(x)=x-1,$ $\forall$ $x\in[1,2)$
and then in 1 the hole will be filled but again, in 2 it will have a hole
and so on
$n=3: [2,3)$ and $F(x)=2x-3$
And i want to establish that
$\lim_{x\to n} F(x)=F(n)$
and by definition $\epsilon$-$\delta$ of continuity i want to make it sure that $F$ is derivable at $x=n$ for $n\in N$
Please can you guys help me with this one? i have/know the idea but as always i dont know how to write it
We claim that $\lim_{x\to n} F(x)=n(n-1)/2=F(n)$.
Given $\epsilon >0$, take $\delta=\frac{\epsilon}{n}$.
If $0<n-x<\delta=\epsilon/n$ then $x<n$, so we use $F(x)$ on the interval $[n-1,n)$ which is $F(x):=(n-1)x-\frac{(n-1)n}{2}$. Then $|F(x)-n(n-1)/2|=|(n-1)x-n(n-1)|=(n-1)(n-x)<(n-1)\cdot \delta=(n-1)\frac{\epsilon}{n}<\epsilon$.
If $0\leq x-n<\delta=\epsilon/n$ we have $x\geq n$ so we use $F(x)$ on the interval $[n,n+1)$ which is $F(x)=nx-n(n+1)/2$.
Now $|F(x)-n(n-1)/2)|=|nx-n^2|=n(x-n)<n\delta=n\cdot \frac{\epsilon}{n}=\epsilon$.