Where does the function $f(x) =\Big[\frac{1}{2}*x\Big]$ contain discontinuities, left or right continuous?

Question

I'm trying to get a grasp on this problem here. I'm going through a calculus textbook to prep myself for a tutoring job. However, i came across this one problem i couldn't seem to make sense of.

$$f(x) =\Big[\frac{1}{2}*x\Big]$$

I eventually decided that it was $continuous$ based on a theorem that stated all polynomials where left and right continuous. However, upon looking at the solution it stated,

"The function is discontinuous at even integers, at which there are jump discontinuities. Because,

$$\lim_{x\to 2n^+} \Big[\frac{1}{2}*x\Big]=n$$ but, $$\lim_{x\to 2n^-} \Big[\frac{1}{2}*x\Big]=n-1$$ it follows that this function is right-continuous at even integers but not left continuous." I still couldn't make sense of it even presented with the answer. I typed this problem into Wolfram Alpha and it said that it was left and right continuous. So how did they decide upon 2n for our limit? How did they get $n-1$ when plugging it into the left sided limit? This problem is making me doubt myself as being an able tutor.

Answer 1

When you type it into wolfram alpha, [] are treated as a form of parenthesis I believe. However, I think in this context: [] means greatest integer function.

So:

floor(1/2*x)

Answer 2

I think that the problem is notational. Square brackets often denote the "floor function" or "integer part function", however Wolfram Alpha does not use this notation, and sees square brackets just as usual brackets (in fact in this case Wolfram Alpha just ignores them).

So, if you interpret $\left[ \frac{1}{2} \cdot x\right]$ as a polynomial the function is continuous, but if you interpret brackets as the floor function, it is not continuous anymore (since its range is the set of integers).