What is $\displaystyle\lim_{x\to\infty}(\left \lfloor{-\dfrac{1}{x}}\right \rfloor )$?

Question

What is $\displaystyle\lim_{x\to\infty}(\left \lfloor{-\dfrac{1}{x}}\right \rfloor )$?

Why is it not -1? Book says 1.

Answer 1

As $x\to+\infty$, $-\frac1x$ approaches $0$ from the negative side; it is negative for any sufficiently large finite $x$. Thus we must conclude that the limit of its floor is $-1$, this being the floor of an arbitrarily small (in magnitude) negative number.

Answer 2

$$\lim_{x \to \infty} \lfloor -\frac{1}{x} \rfloor $$ $$\lim_{x \to \infty} (-\frac{1}{x}-\{-\frac{1}{x}\})$$ $$=0-1=-1$$