Question Number 112370 by Eric002 last updated on 07/Sep/20
$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\frac{{x}−\lfloor{x}\rfloor}{{x}^{\mathrm{2}} } \\ $$$$\lfloor{x}\rfloor\:{is}\:{floor}\:{function} \\ $$
Commented by kaivan.ahmadi last updated on 07/Sep/20
$$\lfloor\mathrm{0}^{+} \rfloor=\mathrm{0} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow^{+} } \:\frac{{x}}{{x}^{\mathrm{2}} }={lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{\mathrm{1}}{{x}}=+\infty \\ $$$$ \\ $$