<s>[INST] 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?
 [/INST] By definition of the floor function
$$x-\frac{1}{x}-1<\left\lfloor x-\frac{1}{x}\right\rfloor \le x-\frac{1}{x}$$
So for $x>0$
$$x^2-1-x < x \left\lfloor x-\frac{1}{x}\right\rfloor \le x^2-1$$
so by the Squeeze Theorem, 
$$\lim_{x\to 0^+}{x \left\lfloor x-\frac{1}{x}\right\rfloor}=-1$$
Now for $x<0$ the direction of the inequality reverses so
$$x^2-1-x > x \left\lfloor x-\frac{1}{x}\right\rfloor \ge x^2-1$$
and again by the Squeeze Theorem, 
$$\lim_{x\to 0^-}{x \left\lfloor x-\frac{1}{x}\right\rfloor}=-1$$
Since the two one-sided limits are the same, the limit as $x\to 0$ is $-1$.
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