Compute $\lim_{x\to\infty} x \lfloor \frac{1}{x} \rfloor$ 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?
 A: 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 $\lim_{x\to\infty}x\lfloor\frac{1}{x}\rfloor=0$.
A: No mistakes. But, can you see it in terms of sequences instead? That will be much easier. $\lim\limits_{x \rightarrow \infty}$ can be viewed as a sequence $\left\lbrace x_n \right\rbrace$, where $\forall M > 0, \exists N \in \mathbb{N}$ such that $x_n > M$. This tells us something special for the choice of $M \geq 1$. Once we choose $M \geq 1$, we will get a stage after which $x_n > 1$ and hence $\left\lfloor{\dfrac{1}{x_n}}\right\rfloor = 0$. Hence, the image sequence is evetually zero and $\lim\limits_{x \rightarrow \infty} x \left\lfloor{\dfrac{1}{x}}\right\rfloor = 0$.
A: The function is $0$ for $x>1$. Alternatively, substitute $m=1/x$. This limit is the same as $\displaystyle\lim_{m\to0^+}\frac{\lfloor m\rfloor}m$, which easily evaluates to $0$.
