Question Number 29156 by abdo imad last updated on 04/Feb/18
![find lim_(x→0^+ ) x[(1/x)] and lim_(x→0^+ ) x^2 [ (1/x)] . [α] is the greatest integr inferior or equal to α.](https://www.tinkutara.com/question/Q29156.png)
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{x}\left[\frac{\mathrm{1}}{{x}}\right]\:\:{and}\:\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:\:{x}^{\mathrm{2}} \:\left[\:\frac{\mathrm{1}}{{x}}\right]\:\:. \\ $$$$\left[\alpha\right]\:{is}\:{the}\:{greatest}\:{integr}\:{inferior}\:{or}\:{equal}\:{to}\:\alpha. \\ $$
Commented by abdo imad last updated on 09/Feb/18
![we have [t]≤t<[t]+1⇒ [(1/x)]≤ (1/x)<[(1/x)]+1 ⇒ ∀x>0 x[ (1/x)]≤1 <x[(1/x)]+x⇒0≤1−x[(1/x)]<x but lim_(x→0^+ ) x=0 ⇒ lim_(x→0^+ ) x[(1/x)]=1.](https://www.tinkutara.com/question/Q29519.png)
$$\:{we}\:{have}\:\:\:\:\left[{t}\right]\leqslant{t}<\left[{t}\right]+\mathrm{1}\Rightarrow\:\:\left[\frac{\mathrm{1}}{{x}}\right]\leqslant\:\frac{\mathrm{1}}{{x}}<\left[\frac{\mathrm{1}}{{x}}\right]+\mathrm{1} \\ $$$$\Rightarrow\:\forall{x}>\mathrm{0}\:\:\:\:{x}\left[\:\frac{\mathrm{1}}{{x}}\right]\leqslant\mathrm{1}\:<{x}\left[\frac{\mathrm{1}}{{x}}\right]+{x}\Rightarrow\mathrm{0}\leqslant\mathrm{1}−{x}\left[\frac{\mathrm{1}}{{x}}\right]<{x} \\ $$$${but}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{x}=\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:{x}\left[\frac{\mathrm{1}}{{x}}\right]=\mathrm{1}. \\ $$
Commented by abdo imad last updated on 09/Feb/18
![for x>0 wehave 0≤ 1−x[(1/x)]<x⇒0≤x −x^2 [(1/x)]<x^2 ⇒ −x≤ −x^2 [(1/x)]< x^2 −x ⇒ x−x^2 < x^2 [(1/x)]≤x but lim_(x→0) (x−x^2 )=lim_(x→0) x =0 ⇒ lim_(x→0) x^2 [(1/x)]=0](https://www.tinkutara.com/question/Q29555.png)
$${for}\:{x}>\mathrm{0}\:{wehave}\:\mathrm{0}\leqslant\:\mathrm{1}−{x}\left[\frac{\mathrm{1}}{{x}}\right]<{x}\Rightarrow\mathrm{0}\leqslant{x}\:−{x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]<{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:−{x}\leqslant\:−{x}^{\mathrm{2}} \:\left[\frac{\mathrm{1}}{{x}}\right]<\:{x}^{\mathrm{2}} −{x}\:\Rightarrow\:{x}−{x}^{\mathrm{2}} <\:{x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]\leqslant{x}\:{but} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left({x}−{x}^{\mathrm{2}} \right)={lim}_{{x}\rightarrow\mathrm{0}} {x}\:=\mathrm{0}\:\Rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}} {x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\right]=\mathrm{0} \\ $$