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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?

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

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  • $\begingroup$ Thank you, I see what you did there. (right click the equation and under "Show Math As" select Tex or MathML ;) $\endgroup$ – user531476 Dec 24 '18 at 13:34
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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 $\lim_{x\to\infty}x\lfloor\frac{1}{x}\rfloor=0$.

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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 $0$.

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

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