Question Number 54688 by pooja24 last updated on 09/Feb/19
$${f}\left({x}\right)=\frac{\mathrm{2}\left[{x}\right]}{\mathrm{3}{x}−\left[{x}\right]}\:{examine}\:{its}\:{continuity}\:{at}\:{x}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${where}\:\left[{x}\right]\:{is}\:{greatest}\:{integer}\:{function} \\ $$
Commented by maxmathsup by imad last updated on 10/Feb/19
$${we}\:{have}\:{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{2}\left(−\mathrm{1}\right)}{−\frac{\mathrm{3}}{\mathrm{2}}−\left(−\mathrm{1}\right)}\:=\frac{−\mathrm{2}}{−\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\:=\frac{−\mathrm{2}}{−\frac{\mathrm{1}}{\mathrm{2}}}\:=\mathrm{4} \\ $$$${and}\:{lim}_{{x}\rightarrow\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{+} } \:\:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{−} } \:\:\:{f}\left({x}\right)=\mathrm{4}\:={f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:{so}\:{f}\:{is}\:{continue}\:{at} \\ $$$${x}_{\mathrm{0}} =−\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Answered by kaivan.ahmadi last updated on 09/Feb/19
$$\mathrm{f}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{li}\underset{\mathrm{x}\rightarrow−\frac{\mathrm{1}^{+} }{\mathrm{2}}} {\mathrm{m}f}\left(\mathrm{x}\right)=\mathrm{li}\underset{\mathrm{x}\rightarrow−\frac{\mathrm{1}^{−} }{\mathrm{2}}} {\mathrm{m}f}\left(\mathrm{x}\right)=\mathrm{4} \\ $$$$\mathrm{so}\:\mathrm{f}\:\mathrm{is}\:\mathrm{continuos}\:\mathrm{at}\:\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$