Solving a sequence limit with floor I've started learning sequences and I'm having a hard time calculating the following, for $a > 0$:
$$\lim_{n\to ∞}{\frac{\lfloor na\rfloor}{n}} $$
Using Heine’s Lemma I'm trying to solve it analogous to the corresponding limit definitions for functions, but I get stuck. I've tried mostly with the Squeeze theorem.
Any help is appreciated.
 A: We know that for any $\alpha$,
$$\alpha-1<[\alpha]\leq \alpha$$
thus:
$$\frac{nx-1}{n}<\frac{[nx]}{n}\leq\frac{nx}{n} $$
Now squeeze to get
$$\lim_{n\to\infty}\frac{[nx]}{n}=x $$
A: Hint. Note that $na-1<\lfloor na\rfloor \leq na $ which implies that
$$a-\frac{1}{n}=\frac{na-1}{n}<\frac{\lfloor na\rfloor}{n} \leq \frac{na}{n}=a\implies \left|\frac{\lfloor na\rfloor}{n}-a\right|<\frac{1}{n}.$$
A: Generally, whenever we encounter greatest integer functions, we create a bound for it and try to apply sandwich theorem for the limit. Here, $na-1\le[na]\le na$
A: Since $\forall n,a : na-1 < \lfloor na \rfloor\leq na$ we can write 
$$
\lfloor na \rfloor na = na - \alpha(n,a)
$$ 
with $\forall a :0 \leq  \alpha(n,a) < 1$.  Now apply $n(\epsilon)-\epsilon$ definition of limit:  Take any given $\epsilon > 0$, and let $n(\epsilon) = \frac2\epsilon$. Then for all $n > n(\epsilon)$, 
$$
a - \frac{\lfloor na \rfloor}{n} = a - \frac{na - \alpha(n,a)}{n} = 
a - \left(a -  \frac{- \alpha(n,a)}{n}\right) = \frac{ \alpha(n,a)}{n} < \frac{ \alpha(n,a)\epsilon}{2} < \epsilon
$$
and also $a - \frac{\lfloor na \rfloor}{n} \geq 0$ so
$$\left| a - \frac{\lfloor na \rfloor}{n} \right| < \epsilon$$
thus showing that the desired limit is $a$.
