Continuity of $f(x) = \left\lfloor {x^2} \right\rfloor - {\left\lfloor {x} \right\rfloor}^2$ The problem asked whether the function in question was continuous for integers.
My attempt:
A function $f$ is continuous at $a$ if $\lim\limits_{x \to a}{f(x)} = f(a)$. For this, I believe that I must first prove that the limit exists, i.e.,
$\lim\limits_{x \to a^-}{f(x)} = \lim\limits_{x \to a^+}{f(x)}$ 
$\equiv \lim\limits_{\epsilon \to 0}{f(n-\epsilon)} = \lim\limits_{\epsilon \to 0}{f(n+\epsilon)}$, for some $n \in \mathbb{I}$
On simplifying, I got:
Left limit = $\lim\limits_{\epsilon \to 0}{\left\lfloor{n^2+\epsilon^2-2n\epsilon}\right\rfloor}-(n-1)^2 = 2n-1$
Right limit = $\lim\limits_{\epsilon \to 0}{\left\lfloor{n^2+\epsilon^2+2n\epsilon}\right\rfloor - n^2}=0$
Which doesn't seem useful....
 A: Let $[.]$ denote GIF and let
$$F(x)=[x^2]-[x]^2$$
$F(x)$ is suspected to be discontinuous at integers $n$. Let  $h$ to be positive and as small as we want. $F(n)=0$.
Next $F(n+h)=[n^2+2hn+h^2]-[(n+h)]^2=n^2+[2nh+h^2]-n^2$. Evenif $n$ us very large we can choose $h$ to much smaller to get $[2nh+h^2]=0$. So $F(n+h)=0.$
Further, $F(n-h)=[n^2-2nh+h^2]-[(n-h)]^2=n^2+[-2nh+h^2]-(n-1)^2=n^2-1-n^2+2n-1=2(n-1).$
For the continuity at $x=n$, we need to have $F(n+h)=F(n)=F(n-h)\implies n=1$
Consequently, among all integers $F(x)$ is continuous only at $x=1$.
A: Let $\;n\in\mathbb Z\;.$
$\begin{align}
&\text{There are three possible cases: }
\,n\leqslant-1\;\text{or}\;n=0\;\text{or}\;n\geqslant1\;.
\end{align}$
First case :$\;n\leqslant-1$
If $\;n\leqslant-1\;,\;$ then it results that
$\lim_\limits{x\to n^-}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=n^2-\left(n-1\right)^2=2n-1\;\;,$
$\lim_\limits{x\to n^+}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=\left(n^2-1\right)-n^2=-1\;\;.$
In order to calculate correctly the previous two limits, we need to pay attention to this important information:
for $\;x\to n^-\;,\;x^2>n^2\;,\;$ hence $\;\lfloor x^2\rfloor=n^2\;\;,$
for $\;x\to n^+\;,\;x^2<n^2\;,\;$ hence $\;\lfloor x^2\rfloor=n^2-1\;\;.$
Second case :$\;n=0$
If $\;n=0\;,\;$ then it results that
$\lim_\limits{x\to n^-}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=0-\left(-1\right)^2=-1\;\;,$
$\lim_\limits{x\to n^+}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=0-0=0\;\;,$
Third case :$\;n\geqslant1$
If $\;n\geqslant1\;,\;$ then it results that
$\lim_\limits{x\to n^-}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=\left(n^2-1\right)-\left(n-1\right)^2=2n-2\;\;,$
$\lim_\limits{x\to n^+}\left(\lfloor x^2\rfloor-\lfloor x\rfloor^2\right)=n^2-n^2=0\;\;.$
If $\;n\leqslant0\;,\;$ the left limit and the right limit are different, consequently the function $\;f(x)=\lfloor x^2\rfloor-\lfloor x\rfloor^2\;$ cannot be continuous at $\;x=n\;$ for all $\;n\leqslant0\;.$
But, if $\;n\geqslant1\,,\,$ the left limit, the right limit and the value of the function $\;f(x)\;$ at $\;x=n\;$ are equal only when $\;2n-2=0\;$ that is $\;n=1\;.$
Hence, the function $\;f(x)=\lfloor x^2\rfloor-\lfloor x\rfloor^2\;$ is not continuous at $\;x=n\;$ for all $\;n\in\mathbb Z\;\land\;n\ne1\;.$
Nevertheless the function $\;f(x)\;$ is continuous at $\;x=1\;.$
A: Taking $(+/-)\epsilon$ and any $n\in \mathbb{Z}$ we get
(in order to have continuity)
that $n^{2}-1-(n-1)^{2}=n^{2}-n^{2}=0$ which gives $2(n-1)=0$ and
hence $n=1$. So I agree with the answer given by Z Ahmed.!
