find-lim-x-0-x-1-x-and-lim-x-0-x-2-1-x-is-the-greatest-integr-inferior-or-equal-to-

Question Number 29156 by abdo imad last updated on 04/Feb/18

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{x}\left[\frac{\mathrm{1}}{{x}}\right]\:\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:{x}^{\mathrm{2}} \:\left[\:\frac{\mathrm{1}}{{x}}\right]\:\:. \\ $$$$\left[\alpha\right]\:{is}\:{the}\:{greatest}\:{integr}\:{inferior}\:{or}\:{equal}\:{to}\:\alpha. \\ $$

Commented by abdo imad last updated on 09/Feb/18

$$\:{we}\:{have}\:\:\:\:\left[{t}\right]\leqslant{t}<\left[{t}\right]+\mathrm{1}\Rightarrow\:\:\left[\frac{\mathrm{1}}{{x}}\right]\leqslant\:\frac{\mathrm{1}}{{x}}<\left[\frac{\mathrm{1}}{{x}}\right]+\mathrm{1} \\ $$$$\Rightarrow\:\forall{x}>\mathrm{0}\:\:\:\:{x}\left[\:\frac{\mathrm{1}}{{x}}\right]\leqslant\mathrm{1}\:<{x}\left[\frac{\mathrm{1}}{{x}}\right]+{x}\Rightarrow\mathrm{0}\leqslant\mathrm{1}−{x}\left[\frac{\mathrm{1}}{{x}}\right]<{x} \\ $$$${but}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{x}=\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:{x}\left[\frac{\mathrm{1}}{{x}}\right]=\mathrm{1}. \\ $$

Commented by abdo imad last updated on 09/Feb/18

$${for}\:{x}>\mathrm{0}\:{wehave}\:\mathrm{0}\leqslant\:\mathrm{1}−{x}\left[\frac{\mathrm{1}}{{x}}\right]<{x}\Rightarrow\mathrm{0}\leqslant{x}\:−{x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]<{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:−{x}\leqslant\:−{x}^{\mathrm{2}} \:\left[\frac{\mathrm{1}}{{x}}\right]<\:{x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}−{x}^{\mathrm{2}} <\:{x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]\leqslant{x}\:{but} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left({x}−{x}^{\mathrm{2}} \right)={lim}_{{x}\rightarrow\mathrm{0}} {x}\:=\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]=\mathrm{0} \\ $$

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