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.
 A: Sorry, to long for a comment, will delete it in the future. Here is exactly what I mean by my comment.
Assume that on some interval $I=(a-b,a+b)$ around $a$ we have.
$$f(x) \leq g(x) \leq h(x)   \, \forall x\in I \backslash \{ a \}$$
and the outside limits are easy, meaning that you can prove with $\epsilon-\delta$ that $\lim_{x \to a} f(x) =\lim_{x \to a} h(x)= L$. Then you get for free a proof with $\epsilon-\delta$ that $\lim_{x \to a} g(x)= L$.
Indeed
$$f(x) \leq g(x) \leq h(x)  \Rightarrow f(x) -L  \leq g(x)-L \leq h(x)-L \Rightarrow $$
$$\left|g(x)-L \right| \leq  \max\{ \left| f(x) -L  \right| ,  \left| h(x) -L  \right| \} (*)$$
Now, pick an $\epsilon >0$, pick the corresponding $\delta_1$ for $g$ and $\delta_2$ for $h$ and set $\delta = \min \{ \delta_1, \delta_2 \}$. Thus
Then if $0 < |x-a | < \delta$ you have $0 < |x-a | < \delta_1$ and $0 < |x-a | < \delta_2$
$$ \left| f(x) -L  \right| < \epsilon,  \left| h(x) -L  \right| < \epsilon \,,$$
and if you plug these in $(*)$ you are done.
For this problem, the heuristic reason why I think that, no matter what the approach is, if it is simple it is a hidden squeeze theorem argument:  the simplest way of relating $\lfloor y \rfloor$ to $y$ for all real $y$ at once is $y-1 \leq \lfloor y \rfloor \leq y$. Moreover, the bounds can be attained, so you can't improve it.  But once you do that, it becomes exactly the argument I included.
A: Hint: $$ \left|x \left\lfloor \frac{1}{x} \right\rfloor - 1\right| = \left|x\left(\left\lfloor \frac{1}{x} \right\rfloor - \frac1x\right)\right| \le |x| = |x-0|$$
