Question Number 79992 by john santu last updated on 29/Jan/20
![lim_(x→0) [(1/x)] = ?](Q79992.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\mathrm{x}}\right]\:=\:? \\ $$
Commented by jagoll last updated on 30/Jan/20
$$\mathrm{do}\:\mathrm{not}\:\mathrm{exis}\: \\ $$
Commented by mathmax by abdo last updated on 30/Jan/20
![x<[x]+1 ⇒[x]>x−1 ⇒[(1/x)]>(1/x)−1 we have lim_(x→0^+ ) ((1/x)−1)=+∞ ⇒lim_(x→0^+ ) [(1/x)] =+∞ for x<0 [(1/x)]≤(1/x) we have lim_(x→0^− ) (1/x)=−∞ ⇒lim_(x→0^− ) [(1/x)]=−∞](Q79996.png)
$${x}<\left[{x}\right]+\mathrm{1}\:\Rightarrow\left[{x}\right]>{x}−\mathrm{1}\:\Rightarrow\left[\frac{\mathrm{1}}{{x}}\right]>\frac{\mathrm{1}}{{x}}−\mathrm{1}\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)=+\infty \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left[\frac{\mathrm{1}}{{x}}\right]\:=+\infty\:\:{for}\:{x}<\mathrm{0}\:\:\left[\frac{\mathrm{1}}{{x}}\right]\leqslant\frac{\mathrm{1}}{{x}}\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\frac{\mathrm{1}}{{x}}=−\infty \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\left[\frac{\mathrm{1}}{{x}}\right]=−\infty \\ $$