Let $$f(x)=\begin{cases}x^2\sum\limits_{r=0}^{\left[\frac{1}{|x|}\right]}r&x\neq0\\k/2&x=0\end{cases}$$ where [.] denotes the greatest integer function. Find the value of $k$ such that $f(x)$ becomes continuous at $x=0$
$\displaystyle x^2\sum\limits_{r=0}^{\left[\frac{1}{|x|}\right]}r=x^2\frac{\left(\left[\frac{1}{|x|}\right]+1\right)\left[\frac{1}{|x|}\right]}{2}$
I found two functions, $g(x)=(1-|x|)/2$ and $h(x)=(1+|x|)/2$ which seem to satisfy the conditions of sandwich theorem for all the values I've checked so far, however I can't think of way to prove this mathematically.
Sandwich theorem states: If $f(x)≤g(x)≤h(x)$ for all numbers, and at some point $x=k$ we have $f(k)=h(k)$, then $g(k)$ must also be equal to them
First of all, we can ignore negative $x$, since $f(-x) = f(x)$ for all $x \in \mathbb R$. So we are interested in the limit $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x^2 \sum_{r=0}^{\lfloor 1/x \rfloor} r = \lim_{x \to 0^+} x^2 \frac{(\lfloor 1/x \rfloor + 1)\lfloor 1/x \rfloor}{2}.$$ Since $$1/x - 1 < \lfloor 1/x \rfloor \le 1/x$$ for all $x > 0$, it follows that $$\frac{1-x}{2} = x^2 \frac{(1/x)(1/x - 1)}{2} < x^2 \frac{(\lfloor 1/x \rfloor + 1)\lfloor 1/x \rfloor}{2} \le x^2 \frac{(1/x + 1)(1/x)}{2} = \frac{1+x}{2}.$$ Now apply the squeeze/sandwich theorem.