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I am trying to evaluate the limit of

$x[1/x]$ as x tends to zero, where $[.]$ is greatest integer function. I know this is dumb question but can I write [1/x] as 1/[x]? Thanks in advance.

Natasha J
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    Note that if you take $x=\frac12$, then $\left\lfloor\frac1x\right\rfloor=\lfloor 2\rfloor=2$, while $\frac1{\lfloor x\rfloor}=\frac10$ isn’t even defined. – Brian M. Scott Mar 29 '21 at 04:00

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Firstly, no question is dumb!

$$\left\lfloor \frac{1}{x}\right\rfloor \ne \frac{1}{\lfloor x\rfloor}$$ in general. For example, take $x = 1.5$ and notice how the LHS and RHS are different. Let's solve the limit you've mentioned, now. $$\lim_{x\to 0} x\left\lfloor \frac{1}{x}\right\rfloor = \lim_{x\to 0} \left( x\cdot \frac{1}{x} - x\left\{ \frac{1}{x}\right\}\right) = \lim_{x\to 0} \left(1 - x\left\{ \frac{1}{x}\right\}\right)$$ Note that $\{x\}$ denotes the fractional part of $x$, and $x = \{x\} + [x]$. Also, $0\le \{x\} < 1$ for all $x$. Hence, $$0\le \left\{ \frac{1}{x}\right\} < 1$$ $$0\le x\left\{ \frac{1}{x}\right\} < x$$ So, $$\lim_{x\to 0} x\left\{ \frac{1}{x}\right\} = 0$$ by the squeeze theorem. This gives $$\lim_{x\to 0} x\left\lfloor \frac{1}{x}\right\rfloor = 1 $$

stoic-santiago
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