$\lim_{x\to0+}(\lfloor x \rfloor + \lfloor1-x\rfloor)$
$\lim_{x\to0-}(\lfloor x \rfloor + \lfloor1-x\rfloor)$
I tried to solve by graphing $(\lfloor x \rfloor + \lfloor1-x\rfloor)$
Graph of the function $(\lfloor x \rfloor + \lfloor1-x\rfloor)$ is straight line $y=0$:
https://www.desmos.com/calculator/la8vnf0z0f
But function doesn't gives '0' for all values of x.
As $(\lfloor 5 \rfloor + \lfloor1-5\rfloor) =1$
Then why graph of function is straight line $y=0$?
Best Answer
Look at the superimposed graphs of $\lfloor x\rfloor$ (blue) and $\lfloor 1-x \rfloor$ (red). Observe that there is a break at every right end point of both functions (because $x,(1-x)$ themselves become an integer).
What you now have is, for, $$\lim_{x\rightarrow0+} \lfloor x\rfloor=0\quad\text{and}\quad\lim_{x\rightarrow0-} \lfloor x\rfloor=-1 \quad\text{but}\quad\lfloor0\rfloor=0$$ $$\lim_{x\rightarrow0+} \lfloor 1-x\rfloor=0\quad\text{and}\quad\lim_{x\rightarrow0-} \lfloor 1-x\rfloor=1\quad\text{but}\quad\lfloor1-0\rfloor=1$$ Hence, $$\lim_{x\rightarrow0+}( \lfloor x\rfloor+\lfloor 1-x\rfloor)=0 \quad\text{and}\quad\lim_{x\rightarrow0-} ( \lfloor x\rfloor+\lfloor 1-x\rfloor)=0\quad\text{but}\quad\lfloor 0\rfloor+\lfloor 1-0\rfloor=1$$
The correct statement to all this is that the limit of the function $\lfloor x\rfloor+\lfloor 1-x\rfloor$ exists at $0$ despite pointwise discontinuity. This is true because of the equality of left-hand and right-hand limits despite the true value of the function being different.