Where is $g(x)=x\lfloor x+1 \rfloor$ continuous?

Question

Determine the points of continuity of $$g(x)=x\lfloor x+1 \rfloor$$

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I know the point of continuity of $f(x)=\lfloor x+1 \rfloor$ are $x \in (a,a+1)$ where $a \in \mathbb{Z}$. But how do I determine the points of continuity of $x\lfloor x+1 \rfloor$? One idea I have is to consider the graph of $x(x+1)$ but I'm not sure how to proceed.

Answer 1

Case.(1) $x\in [n+\delta,n+1-\delta]$, $g(x)=x\cdot(n+1)$, so $g$ is continous inside each invertal.

Case.(2) at each integer point $n$,

$$\lim_{x\to n^+}g(x)=n(n+1),~~~\lim_{x\to n^-}g(x)=n\cdot n$$

therefore,

$$\lim_{x\to n^+}g(x)=\lim_{x\to n^-}g(x)\Longleftrightarrow n=0$$

Namely, $g$ is continuous at $n=0$, and discontinuous at other integer points $n\neq0$

Here is a plot for function $g$

Answer 2

Let $\;a\in\Bbb Z\,.$

Since $\;\lfloor x+1\rfloor=\begin{cases}a&\text{for any }\;x\in[a-1,a)\\a+1&\text{for any }\;x\in[a,a+1)\quad,\end{cases}$

it follows that

$\lim\limits_{x\to a^-}g(x)=\lim\limits_{x\to a^-} x\lfloor x+1\rfloor=a^2\;\;$ and

$\lim\limits_{x\to a^-}g(x)=\lim\limits_{x\to a^-} x\lfloor x+1\rfloor=a(a+1)=g(a)\;.$

On the other hand,

$a^2=a(a+1)\;$ if only if $\;a=0\;,\;$ consequently,

the only point of continuity of the function $\,g(x)\,$ which belongs to $\,\Bbb Z\,,\;$ is $\,a=0\,.$

Moreover, in any point $\;x_0\in\Bbb R\setminus\Bbb Z\,,\,$ the function $\,g(x)\,$ is continuous because both $\,\varphi(x)=x\,$ and $\,\psi(x)=\lfloor x+1\rfloor\,$ are continuous in $\,x_0\,,$ indeed $\;x_0\in\big(\lfloor x_0\rfloor,\lfloor x_0\rfloor+1\big)$ and $\;\psi(x)=\lfloor x+1\rfloor=\lfloor x_0\rfloor+1\;$ is a constant function on the interval $\big(\lfloor x_0\rfloor,\lfloor x_0\rfloor+1\big).$

Hence the function $\,g(x)\,$ is continuous on $\big(\Bbb R\setminus\Bbb Z\big)\cup\big\{0\big\}.$