Limit, Greatest Integer function?

Question

Q. Find $\lim _{x\to 0}\left(1-x+\left[x-1\right]+\left[1-x\right]\right)$ where $\left[y\right]$ denotes the greatest integer function not exceeding 'y'.

Answer 1

If $L$ us the value of your limit, then using the fact that $[x+n]=[x]+n$ for all $x\in\mathbb R, n\in\mathbb Z$, you'll have $$L=\lim_{x\to 0}(1-x+[x]-1+1+[-x])=\lim_{x\to 0}(1-x+[x]+[-x])=\lim_{x\to 0}(1-x-1)=0.$$

Answer 2

Try putting $x=0^+$ (meaning a bit larger than zero) and $x=0^-$ (a bit smaller than zero) and see what happens.

in the first case, $-1<x-1<0$ and so $[x-1]=-1$ and with a similair argument, $[1-x]=0$

And so in the first example, if $x$ is a bit larger than $0$, the result is $0$.

Do the same thing when $x=0^-$ to see if it is the same answer. if it is, then the answer is $0$. if it isn't, then there is no limit.

Answer 3

We use the fact that

$\lfloor x \rfloor +\lfloor -x\rfloor = \left\{ \begin{array}{lr} 0 & : x \in \mathbb{Z}\\ -1 & : x \not\in \mathbb{Z} \end{array} \right. $

As $x$ becomes exactly $0$, the expression is $1$.

But when $x$ approaches $0$, $x$ still cannot become an integer, which gives an expression $0$.

Therefore, the limit doesn't exist. ~~~