# 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.

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 $$\lim_{n\to\infty}\frac{[nx]}{n}=x$$
Generally, whenever we encounter greatest integer functions, we create a bound for it and try to apply sandwich theorem for the limit. Here, $na-1\le[na]\le na$
Hint. Note that $na-1<\lfloor na\rfloor \leq na$ which implies that $$a-\frac{1}{n}=\frac{na-1}{n}<\frac{\lfloor na\rfloor}{n} \leq \frac{na}{n}=a\implies \left|\frac{\lfloor na\rfloor}{n}-a\right|<\frac{1}{n}.$$
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 $a$.