Let $\theta>1$ and prove that $\lim_{m \to \infty} \frac{\lfloor\theta(2^m-m)\rfloor}{2^m-m} = \theta$ 
Let $\theta>1$ and prove that $\displaystyle \lim_{m \to \infty} \dfrac{\lfloor\theta(2^m-m)\rfloor}{2^m-m} = \theta$.

I was wondering how to deal with the floor function in order to prove this limit.
 A: You just need to use $m-1<\lfloor m\rfloor\leq m$.
$\lim_{m \to \infty} \dfrac{\lfloor\theta(2^m-m)\rfloor}{2^m-m} = \lim_{m \to \infty} \dfrac{\theta(2^m-m)-\delta}{2^m-m}$ where $0\leq \delta<1$ and $2^m-m\to \infty$ as $m$ goes to infinity.
A: write $n$ for $2^m-m$
$$
 n\theta \ge \lfloor{n\theta}\rfloor \ge n\theta -1
$$
so
$$
\theta \ge \frac{\lfloor{n\theta}\rfloor}{n} \ge \theta - \frac1{n}
$$
A: $2^m-m$ is an integer, so if you can prove that $\lim\limits_{n\to\infty} \dfrac{\lfloor \theta n\rfloor} n =\theta,$ with $n$ running through all integers, then the limit you're asking about follows because if a sequence converges then every one of its subsequences converges to the same limit.
You have $\theta n-1 \le \lfloor\theta n\rfloor \le \theta n$, so
$$
\theta - \frac 1 n = \frac{\theta n - 1} n \le \frac{\lfloor \theta n\rfloor} n \le \frac{\theta n} n = \theta.
$$
If $\lim\limits_{n\to\infty} \left( \theta- \dfrac 1 n \right) =\theta$ and $\lim\limits_{n\to\infty} \theta =\theta,$ then you've got the result by squeezing.
