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Question Number 105036 by bobhans last updated on 25/Jul/20

$$\underset{x\to 0}{\lim }x.\left[\frac{1}{x}\right]?$$$$note\left[\right]=greatestintegerfunction$$

Answered by john santu last updated on 25/Jul/20

$$thelimitasx\to 0^{+}canbe$$$$transformedinto\underset{p\to \mathrm{\infty }}{\lim }\frac{\lfloor p\rfloor }{p}$$$$set\left\{p\right\}=p-\lfloor p\rfloor$$$$wehave\frac{\lfloor p\rfloor }{p}=1-\frac{\left\{p\right\}}{p},since$$$$0⩽\left\{p\right\}<1.so\underset{p\to \mathrm{\infty }}{\lim }(1-\frac{\left\{p\right\}}{p})=$$$$1-0=1.similarlyfor$$$$thelimitasx\to 0^{-}$$$$\left(JS\mathrm{♠}⧫\right)$$

Answered by mathmax by abdo last updated on 25/Jul/20

$$\mathrm{we}\mathrm{have}\left[\frac{1}{\mathrm{x}}\right]⩽\frac{1}{\mathrm{x}}<\left[\frac{1}{\mathrm{x}}\right]+1\Rightarrow \mathrm{for}\mathrm{x}>0\Rightarrow \mathrm{x}\left[\frac{1}{\mathrm{x}}\right]⩽1<\mathrm{x}\left[\frac{1}{\mathrm{x}}\right]+\mathrm{x}\Rightarrow$$$$\{\begin{array}{l}\mathrm{x}\left[\frac{1}{\mathrm{x}}\right]⩽1\Rightarrow 1-\mathrm{x}<\mathrm{x}\left[\frac{1}{\mathrm{x}}\right]⩽1\mathrm{we}\mathrm{passe}\mathrm{to}\mathrm{limit}(\mathrm{x}\to {\mathrm{o}}^{+})\Rightarrow \\ 1-\mathrm{x}<\mathrm{x}\left[\frac{1}{\mathrm{x}}\right]\end{array}$$$${\lim }_{\mathrm{x}\to 0^{+}}\mathrm{x}\left[\frac{1}{\mathrm{x}}\right]=1$$

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