Compute $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ Let $f: \mathbb{R} \to \mathbb{R}$ be the following function:
$f(x) = x \lfloor x - \frac{1}{x} \rfloor$
Show that $f(x)$ admits a limit at zero, and compute its value.

Using epsilon delta, I can prove that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor = - 1$:
$|x \lfloor x - \frac{1}{x} \rfloor + 1|$
$\leq |x \lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |\lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |x| + |1|$
$\leq |x|^2 + |1| < \epsilon$ if $|x| < \delta = \sqrt{\epsilon - 1}$
Is that right? If not, how can I do prove it?
But how do I show that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ exists?

Suppose that $\frac1{n+1}\le x<\frac1n$; then $n<\frac1x\le n+1$, so 
$$-n-\frac{n}{n+1}=\frac1{n+1}-(n+1)\le x-\frac1x<\frac1n-n=-n+\frac1n\;,$$
and $\left\lfloor x-\frac1x\right\rfloor$ is either $-n-1$ or $-n$. Thus, $x\left\lfloor x-\frac1x\right\rfloor$ is either $-nx$ or $-(n+1)x$. Now use the limits on $x$ to bound $-nx$ and $-(n+1)x$, and let $n$ increase without bound.
