Determine where this function is continuous: $$f(x)= \begin{cases} \frac{1}{x-\lfloor x\rfloor}, & \text{for } x \notin \mathbb{Z}\\ 1, & \text{for } x \in \mathbb{Z}\\ \end{cases} $$
Thanks! What I have so far:
Continuity is where the $\displaystyle{\lim_{x \to a}} f(x)$ = $f(a)$.
This is certainly true because if we see $a$ as equal to $n + \delta$, then $$\frac{1}{n + \delta-\lfloor{x + \delta}\rfloor}$$ $$= \frac{1}{n+\delta -n}$$ $$= \frac{1}{\delta}$$
$f(a)$ is also equal to $\frac {1}{\delta}$, so therefore the function is continuous when $\delta \neq 0$. $\delta$ equals $0$ when $a$ is an integer. Therefore, this is continuous in the interval $(x, x+1)$.
Is there something wrong with this line of thought? And if there isn't, is there a more rigorous or faster way of doing this problem? Thanks!
If $a \in \mathbb{Z}$, we have $a+\frac1{n+1} \to a$ but $$f\left( a+\frac1{n+1}\right)=\frac{1}{a+\frac1{n+1}-a}=n+1 \to \infty \ne 1 = f(a)$$
Hence it is not continuous at integer values.
Notice that $f(x)$ is just $\frac1x$ on $(0,1)$, hence it is continuous on $(0,1)$.
Also $f$ is a periodic function with period $1$.
Hence $f$ is continous at $x$ if only if $x$ is not an integer.