In which points the limit of function $f(x)=\lfloor{x^2-1}\rfloor+4x\lfloor x\rfloor+x\lfloor {4x-1}\rfloor$ does not exisct?

Question

For $x\in[-1,1]$, in which points the limit of function $f(x)=\lfloor{x^2-1}\rfloor+4x\lfloor x\rfloor+x\lfloor {4x-1}\rfloor$ doesn't exist?

To solve this problem I think I should find the left-hand side and right-hand side limits for the points $x=0,\frac14,\pm1$ (because they are the roots of the expressions inside the floor functions) and if for example for the point $x=a$ we have $\lim f(x)_{x\to a^+}\neq\lim f(x)_{x\to a^-}$ then the limit doesn't exist at that point.

Is my approach correct ?

Answer 1

Consider $[x^2 - 1]$. If $x = \pm 1$ then $x^2 - 1= 0$ and $[x^2 - 1] = 0$. But if $-1 < x < 1$ then $0 \le x^2 < 1$ and $-1 \le x^2 - 1 < 0$ and $[x^2 - 1]=-1$.

So $[x^2 -1]$ can only have discontinuities (and does have discontinuities) at $x = \pm 1$.

Consider $4x[x]$. If $1\le x < 0$ then $[x]=-1$ and $4x[x] = -4x$. And if $0\le x< 1$ then $[x] = 0$ and $4x[x] = 0$. And if $x=1$ then $[x]=1$ and $4x[x]= 4x$.

So $4x[x]$ can only have disontinuities at $x = 0$ or $x = 1$. $\lim_{x\to 0^-}4x[x]=\lim_{x\to 0} -4x = 0$ and $\lim_{x\to 0^+} 4x[x] =\lim_{x\to 0} 0 = 0$. So there is not discontinuity at $x = 0$. However $\lim_{x\to 1} 4x[x] = \lim_{x\to 1} 0 = 0$ while $4\cdot 1[1] = 4$ so there is a disontinuity at $x = 1$.

Lastly consider $x[4x-1]$. If $\frac n4 \le x < \frac {n+1}4$ (for some integer $n$) then $[4x-1]= n-1$ and and $x[4x-1] = x(n-1)$ and can only have discontinuities at $x = \frac n4$.

So $f(x)$ can only have discontinuities as $x = \frac n4$.

At $x' = -1$ we have $f(x)= 0 + 4 +5 = 9$ while $\lim_{x\to -1^+} f(x) =\lim_{x\to -1^+} -1 -4x -5x = 8$. There is a discontinuity for $x=-1$.

For $x' = \frac n4$ for $n=--3,-2,-1$

We have $\lim_{x\to x'^-} = -1 -4x +(n-2)x$ while $\lim_{x\to x'^+} =-1 -4x + (n-1)x$ which are not equal.

For $x'= 0$ we have $\lim_{x\to 0^-}= -1-4x -x = -1$ and $\lim{x\to 0^+}=-1 +0 + 0=-1$ so there is no discontinuity.

For $x' = \frac n4$ for $n=1,2,3$ we have $\lim_{x\to x'^-} = -1 +(n-2)x$ while $\lim_{x\to x'^+} =-1 + (n-1)x$ which are not equal.

And finally for $x'=1$ we have $f(1)=0 + 4 + 3=7$ where $\lim_{x\to 1}f(x) = \lim_{x\to 1}= 0 + 0 + 2x = 2$. A discontinuity.

SO discontinuities exist at all $x = \frac n4; n\ne 0; n\in \mathbb Z$.