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What would be the limit of x[1/x] as x goes to 0. Where [ ] denote the greatest integer function

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    You think it is $1$, right? Now try an epsilon-delta argument with the limit $1$ and see what goes wrong (if anything). – 2017-01-26
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    Here is another answer, a la Euler: http://mathforum.org/library/drmath/view/69917.html – 2017-01-28
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    Of course, if you simplify it to x/x then the limit is 0/0 which is 1, by definition. – 2017-01-28

2 Answers 2

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$[1/x]\le 1/x<[1/x]+1$ from where you get $1/x-1<[1/x]\le 1/x$

If $x>0$ then by multiplying with $x$ one gets:

$1-x

Similarly, when $x<0$ multiply with $x$:

$1\le x[1/x]< 1-x$, and by taking $x\rightarrow 0_-$ one gets $\lim_{x\rightarrow 0_-}x[1/x]=1$

So your limit is $1$

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Another observation :$$\lim_{x\to 0}x\lfloor\dfrac1x\rfloor$$ take $\dfrac1x=y \\x\to 0 \Rightarrow y\to \infty$ $$\lim_{x\to 0}x\lfloor\dfrac1x\rfloor=\lim_{y\to \infty}\dfrac1y\lfloor y \rfloor=\\ \lim_{y\to \infty}\dfrac{\lfloor y \rfloor}{y}$$ now $$y=n+p\\n\to \infty \\0\leq p<1$$ so $$\lim_{y\to \infty}\dfrac{\lfloor y \rfloor}{y}=\lim_{n\to \infty}\dfrac{n}{n+p}=\\ \lim_{n\to \infty}\dfrac{n}{n(1+\dfrac pn)}=\\\lim_{n\to \infty}\dfrac{1}{(1+\dfrac pn)}=1$$