# 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?

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 $$\lim\limits_{x \rightarrow \infty} x \left\lfloor{\dfrac{1}{x}}\right\rfloor = 0$$.

• Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;) – user531476 Dec 24 '18 at 13:34

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 $$\lim_{x\to\infty}x\lfloor\frac{1}{x}\rfloor=0$$.

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 $$0$$.

• How does it easily evaluate to 0? – user531476 Dec 24 '18 at 13:39
• $\displaystyle\lim_{m\to0^+}\frac{\lfloor m\rfloor}m=\displaystyle\lim_{m\to0^+}\frac{0}m=0$. This is due to the fact that $\lfloor m\rfloor$ is exactly $0$, but $m$ only tends to $0$ – Shubham Johri Dec 24 '18 at 13:41
• Oh, I thought we take that as $\frac{0}{0}$. Thank you! – user531476 Dec 24 '18 at 14:28