$\lfloor\frac{1}{x}\rfloor\cdot x^2$ continuous at $0$

Question

I am trying to prove, that this function k is: 1)continuous at $0$ and 2) noncontinuous in all $x_0=\frac{1}{n}, n\in\{\Bbb Z\}\setminus\{0\} $ the function k is given through: $$ k:== \begin{cases}(\lfloor\frac{1}{x}\rfloor\cdot x^2) &\mbox{if } x\neq0 \\ 0 & \mbox{if } x=0 \end{cases} $$ $$k: \Bbb R\to \Bbb R $$

So, considering the first question I got:

given $ |x-x_0|<\delta$ and since we are working in $0$, $x_0=0$, so $|x|<\delta$

Now I am starting with $$|k(x)-k(x_0)|=\left|\lfloor\frac{1}{x}\rfloor\cdot x^2-0\right|=\left|\lfloor\frac{1}{x}\rfloor\cdot x^2\right|<\left|\lfloor\frac{1}{x}\rfloor\delta^2\right|=\left|\lfloor\frac{1}{x}\rfloor\right|\cdot\delta^2$$ But at this point I do not know what to do with the floor function and I have no idea at the moment how to prove, that it is noncontinuous at $x_0=\frac{1}{n}$

Answer 1

Hint for the first question : You have for all $x \in \mathbb{R^*}$ : $\frac{1}{x}-1 < \left \lfloor\frac{1}{x} \right \rfloor \leq \frac{1}{x}$

Answer 2

For the first question:

Start by noting that $y-1 \le \lfloor y \rfloor \le y$. This leads to

$$ \left(\frac{1}{x} - 1 \right) x^2 \le k(x) \le x,$$

and the squeeze theorem does the job for $0$.

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For the second question:

$$x \in \left(\frac{1}{n+1}, \frac{1}{n}\right) \implies \left\lfloor \frac{1}{x} \right\rfloor = n\\ \implies \lim_{x\to (1/n)^-} k(x) = \lim_{x\to (1/n)^-} n x^2 = \frac{1}{n}$$

$$x \in \left(\frac{1}{n}, \frac{1}{n-1}\right) \implies \left\lfloor \frac{1}{x} \right\rfloor = n-1\\ \implies \lim_{x\to (1/n)^+} k(x) = \frac{n-1}{n^2}$$

Thus the limit doesn't exist at $1/n$ for any integer $n$, and the function is not continuous.

Answer 3

Just want to provide an $\varepsilon$-proof for your reference:

We claim that the map is continuous from the right and the left at $0$, separately.

If $x > 0$, then $$ \bigg| \bigg\lfloor \frac{1}{x} \bigg\rfloor x^{2} \bigg| \leq \frac{1}{x}x^{2} = x; $$ hence, given any $\varepsilon > 0$, if $0 < x < \varepsilon$ then $$ \bigg| \bigg\lfloor \frac{1}{x} \bigg\rfloor x^{2} \bigg| < \varepsilon. $$ We proved the right continuity.

If $x < 0$, then $$ \bigg| \bigg\lfloor \frac{1}{x} \bigg\rfloor x^{2} \bigg| \leq \bigg( 1 - \frac{1}{x} \bigg)x^{2} = x^{2}-x = x(x-1); $$ if in addition $x > -1$, then $x(x-1) \leq -2x$; given any $\varepsilon > 0$, we have $-2x < \varepsilon$ if in addition $x > -\varepsilon/2$. In conclusion, for every $\varepsilon > 0$, if $\max \{ -1, -\varepsilon/2 \} < x < 0$ then $$ \bigg| \bigg\lfloor \frac{1}{x} \bigg\rfloor \bigg| < \varepsilon; $$ we proved the left continuity.