Inspired by Yashasvi Baweja

It is not hard to see that the second limit

$\left\lfloor \lim_{x \rightarrow \infty} \frac{x}{x+1} \right\rfloor = \lfloor 1 \rfloor = 1$

The tricky part is the first limit. Note that $\frac{x}{x+1} < 1$ for all $x \in (-1, \infty)$ (duh) and hence $\left\lfloor \frac{x}{x+1}\right\rfloor = 0$ in that domain. This implies that

$\lim_{x \rightarrow \infty} \left\lfloor \frac{x}{x+1} \right\rfloor = 0$

Hence,

$\lim_{x \rightarrow \infty} \left\lfloor \frac{x}{x+1} \right\rfloor-\left\lfloor \lim_{x \rightarrow \infty} \frac{x}{x+1} \right\rfloor = 0-1 = \boxed{-1}$

This is simple because the floor function is not continuous on any interval, so $\lim_{x \to \infty} f(x) \neq f(\lim_{x \to \infty} x)$ . Therefore, you need to solve this in the order the problem is given.