I'm working out a limit and I'm not sure if my assumption is considered rigorous
$$\lim_{x\to\infty} x\left\lfloor\frac1x\right\rfloor$$
I supposed that $0\leq x\left\lfloor\frac1x\right\rfloor \leq \left\lfloor\frac1x\right\rfloor$ since $x$ is approaching $\infty$ $($thus $x > 1$$)$ and to get the answer $0$.
Any mistakes here?
Best Answer
You make it sound like the reason that $x\lfloor\frac{1}{x}\rfloor\leq\lfloor\frac{1}{x}\rfloor$ is true for positive $x$ is because $xy\leq y$ for positive $x$ and $y$, but this is not so. Instead, the reason it is true is because if $x\leq 1$ then $xy\leq y$ for positive $y$ and if $x>1$ then $\lfloor\frac{1}{x}\rfloor=0$ so that $x\lfloor\frac{1}{x}\rfloor=0$ too.
It is far simpler just to note that if $x>1$ then $\lfloor\frac{1}{x}\rfloor=0$ and hence $x\lfloor\frac{1}{x}\rfloor=0$, giving us $\lim_{x\to\infty}x\lfloor\frac{1}{x}\rfloor=0$.