Vtyjkyuk

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 $F(x)=0$, $forall$ $xin[0,1)$

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 $F(x)=x-1,$ $forall$ $xin[1,2)$

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

$lim_xto n F(x)=F(n)$

and by definition $epsilon$-$delta$ of continuity i want to make it sure that $F$ is derivable at $x=n$ for $nin N$

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 $lim_xto n F(x)=n(n-1)/2=F(n)$.

Given $epsilon >0$, take $delta=fracepsilonn$.

If $0<n-x<delta=epsilon/n$ then $x<n$, so we use $F(x)$ on the interval $[n-1,n)$ which is $F(x):=(n-1)x-frac(n-1)n2$. Then $|F(x)-n(n-1)/2|=|(n-1)x-n(n-1)|=(n-1)(n-x)<(n-1)cdot delta=(n-1)fracepsilonn<epsilon$.

If $0leq x-n<delta=epsilon/n$ we have $xgeq n$ so we use $F(x)$ on the interval $[n,n+1)$ which is $F(x)=nx-n(n+1)/2$.

Now $|F(x)-n(n-1)/2)|=|nx-n^2|=n(x-n)<ndelta=ncdot fracepsilonn=epsilon$.