> 
> Draw the graph and study the discontinuity points of the function
> 
> 
> $$
> h(x)= \begin{cases}
> x[\frac{1}{x}] & \text{if } x \neq 0, \\
>  1 & \text{if } x=0. \\
> \end{cases}
> $$
> 
> 
> 

Where $[\frac{1}{x}]$ is the floor of $\frac{1}{x}$.

My attempt:

[![enter image description here](https://i.stack.imgur.com/HJClX.jpg)](https://i.stack.imgur.com/HJClX.jpg)

* The limit doesn’t exist at $x=1$

$\lim\_{x \to 1^+} h(x)=0$ and $\lim\_{x \to 1^-} h(x)=1 $

Thus, $h$ is not continuous at $x=1$

* $\lim\_{x\to 0^+} h(x)=\ lim\_{x\to 0^-} h(x)=h(0)=1 $

Hence, $h$ is continuous at $x=0$.

The function is continuous on $\Bbb{R}\setminus \{1\}$. Is that true?

Edit:

I noted that

* If $\frac{1}{2}<x\le 1$ then $x[\frac{1}{x}]=x$
* If $\frac{1}{3}<x\le \frac{1}{2}$ then $x[\frac{1}{x}]=2x$
* If $\frac{1}{4} < x \le \frac{1}{3}$ then $x[\frac{1}{x}]=3x$

.

.

* If $\frac{1}{n+1} < x \le \frac{1}{n}$, then $x[\frac{1}{x}]=nx$

And we have that

$\lim\_{x \to \frac{1}{2}^-} f(x)=1$ and $\lim\_{x \to \frac{1}{2}^+} f(x)=\frac{1}{2}$

Then the limit does not exit at $x=\frac{1}{2}$

$\lim\_{x \to \frac{1}{3}^-} f(x)=1$ and $\lim\_{x \to \frac{1}{3}^+} f(x)=\frac{2}{3}$

Then the limit does not exit at $x=\frac{1}{3}$.

In general, $\lim\_{x \to \frac{1}{n}^-}=1\neq \lim\_{x \to \frac{1}{n}^+}=\frac{n-1}{n}$

Then, $f$ is not continuous at $x=\frac{1}{n}$ for all $n\in \Bbb{N}$.