Question Number 85868 by M±th+et£s last updated on 25/Mar/20
$${if}\:{f}\left({x}\right)=\lfloor{x}^{\mathrm{2}} \rfloor\:\: \\ $$$${and}\:{A}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({f}\left({x}\right)−{f}\left(−{x}\right)\right) \\ $$$${and}\:{B}={f}\left({x}\right)+{f}\left(−{x}\right)\:\:{when}\:{x}=\mathrm{0} \\ $$$$ \\ $$$${find}\:{A}\:{and}\:{B} \\ $$
Commented by mr W last updated on 25/Mar/20
$${A}=\mathrm{0}−\mathrm{0}=\mathrm{0} \\ $$$${B}=\mathrm{0}+\mathrm{0}=\mathrm{0} \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
$${thank}\:{you}\:{sir}\:{sir}\:{but}\:{why}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({f}\left({x}\right)−{f}\left(−{x}\right)\right)=\mathrm{0}\: \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
$${sorry}\:{i}\:\:{mean}\:{f}\left({x}\right)+{f}\left(−{x}\right) \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
$$\lfloor{x}\rfloor+\lfloor−{x}\rfloor=\begin{cases}{\mathrm{0}\:\:\:\:{x}\in{z}}\\{−\mathrm{1}\:\:{x}\notin{z}}\end{cases} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\:\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {{lim}}\:\left(−\mathrm{1}\right)=−\mathrm{1} \\ $$$$ \\ $$$${so}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}f}\left({x}\right)+{f}\left(−{x}\right)=−\mathrm{1} \\ $$
Commented by mr W last updated on 25/Mar/20
$${but}\:{here}\:{f}\left({x}\right)=\lfloor{x}^{\mathrm{2}} \rfloor\:{as}\:{you}\:{gave}. \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
$${you}\:{are}\:{right}\:{sir}\:{iam}\:{so}\:{sorry}\:{it}\:{was}\:{a}\:{typo} \\ $$