# 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?

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 ?

• $f(x)=\lfloor x\rfloor$ has a discontinuity at every integer, doesn't it? Apr 12, 2021 at 15:54
• @saulspatz Oh I missed that! Apr 12, 2021 at 15:55

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 $$4\cdot 1 = 4$$ 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 $$x = \frac n4; n\ne 0; n\in \mathbb Z$$.