Use $\epsilon-\delta$ definition of limit to prove that $\displaystyle \lim_{x \to 0} x \lfloor \frac{1}{x} \rfloor = 1$. I was trying to write some nice problems for applying $\epsilon-\delta$ definition to give it to my friend but then I realized that I couldn't solve some of them either. This is one of them: 
Use $\epsilon-\delta$ definition of limit to prove that
$$ \lim_{x \to 0} x \lfloor \frac{1}{x} \rfloor = 1$$
It's easy to show that this is true by using the squeeze(sandwich) theorem, but I'm looking for an $\epsilon-\delta$ proof.
Also, a similar problem could be:
$$ \lim_{n \to \infty} \frac{[nx]}{n}=x$$
Again it's obvious that this is true by using the squeeze theorem, but I'm looking for an elementary proof that uses nothing but just the definition of the limit of a sequence.

Hint: $$ \left|x \left\lfloor \frac{1}{x} \right\rfloor - 1\right| = \left|x\left(\left\lfloor \frac{1}{x} \right\rfloor - \frac1x\right)\right| \le |x| = |x-0|$$
