Graphing $(\lfloor x \rfloor + \lfloor1-x\rfloor)$

Question

$\lim_{x\to0+}(\lfloor x \rfloor + \lfloor1-x\rfloor)$

$\lim_{x\to0-}(\lfloor x \rfloor + \lfloor1-x\rfloor)$

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 $y=0$?

Answer 1

For any $x\in\mathbb Z$: $\lfloor x\rfloor=x$ and $\lfloor 1-x\rfloor=1-x$ so: $\lfloor x\rfloor+\lfloor 1-x\rfloor=x+(1-x)=1$ . However the floor function is discontinuous hence why when evaluating individually the upper and lower limits will be different- the definition of discontinuous.

Answer 2

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.

Answer 3

It is not just showing the $y=0$ line. I beleive you agree that the value of that function is either $0$ or $1$, later occuring when $x$ is an integer.

If you hover your fingers or mouse, whatever you're using, over any integral point then is shows that the range is not just $0$. For example go to $x=5$ the point displayed will be $(5,1)$

The thing is cardinality of real numbers is greater than the cardinality of integers that's why it's not feasible to show those tiny points with different colors.

Answer 4

To calculate $\lim_{x\to 0^+}(\lfloor x \rfloor + \lfloor1-x\rfloor)$, WLOG take $x\in(0,1)$. Then, $1-x\in (0,1)$, so $$\lim_{x\to0^+}(\lfloor x \rfloor + \lfloor1-x\rfloor) = \lim_{x\to0^+}(0 + 0) = 0.$$

To calculate $\lim_{x\to 0^-}(\lfloor x \rfloor + \lfloor1-x\rfloor)$, WLOG take $x\in(-1,0)$. Then, $1-x\in (1,2)$, so $$\lim_{x\to0^+}(\lfloor x \rfloor + \lfloor1-x\rfloor) = \lim_{x\to0^+}(-1 + 1) = 0.$$

Therefore, the limit $\lim_{x\to 0}(\lfloor x \rfloor + \lfloor1-x\rfloor)$ exists and is equal to $0$.

To better understand what is going on, let $x\in (n,n+1)$ for some integer $n$. Then, $1-x \in (-n,1-n)$, so $\lfloor x \rfloor + \lfloor1-x\rfloor = n +(-n) = 0$. If $x = n$ for some integer $n$, then $\lfloor x \rfloor + \lfloor1-x\rfloor = n + (1-n) = 1$. We conclude that

$$\lfloor x \rfloor + \lfloor1-x\rfloor= \begin{cases} 0,& x\not\in\mathbb Z\\ 1,& x\in \mathbb Z\end{cases}$$

so the graph of the function almost is the line $y = 0$, it simply has discontinuities at integer points. If you replaced values of $\lfloor x \rfloor + \lfloor1-x\rfloor$ at integer points with $\lim_{x\to 0}(\lfloor x \rfloor + \lfloor1-x\rfloor)$, you would get continuous function $f(x) = 0$, which graph corresponds to what Desmos plots.