# Old-question-related-to-greatest-int-function-lim-x-0-1-x-1-1-1-lim-x-0-1-x-

Question Number 2624 by prakash jain last updated on 23/Nov/15
$$\mathrm{Old}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{greatest}\:\mathrm{int}\:\mathrm{function}. \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}+{x}\rfloor=\mathrm{1} \\$$$$\lfloor\mathrm{1}\rfloor=\mathrm{1} \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=? \\$$
Answered by Filup last updated on 24/Nov/15
$$\mathrm{if}\:\lfloor{a}\pm{b}\rfloor=\lfloor{a}\rfloor\pm\lfloor{b}\rfloor:\:\:\:\:\:\mathrm{for}\:\:\left({a},\:{b}\right)\in\mathbb{Z} \\$$$$\\$$$$\left.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}+{x}\right]=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}\rfloor+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor \\$$$$=\mathrm{1}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor=\mathrm{1}+\lfloor\mathrm{0}\rfloor=\mathrm{1} \\$$$$\\$$$$\boldsymbol{\mathrm{Similarly}} \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}\rfloor−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor \\$$$$=\mathrm{1}−\lfloor\mathrm{0}\rfloor=\mathrm{1} \\$$
Commented by Yozzi last updated on 24/Nov/15
$${Let}\:{a}=\mathrm{1}.\mathrm{99},{b}=\mathrm{1}.\mathrm{01} \\$$$$\Rightarrow\lfloor\mathrm{1}.\mathrm{99}+\mathrm{1}.\mathrm{01}\rfloor=\lfloor\mathrm{3}\rfloor=\mathrm{3}\neq\lfloor\mathrm{1}.\mathrm{99}\rfloor+\lfloor\mathrm{1}.\mathrm{01}\rfloor=\mathrm{1}+\mathrm{1}=\mathrm{2} \\$$$$\\$$
Commented by Filup last updated on 24/Nov/15
$${T}\mathrm{hanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{proof}! \\$$$${B}\mathrm{ack}\:\mathrm{to}\:\mathrm{the}\:\mathrm{drawing}\:\mathrm{board}! \\$$
Commented by Filup last updated on 24/Nov/15
$${if}\:\:\lfloor{a}\pm{b}\rfloor=\lfloor{a}\rfloor\pm\lfloor{b}\rfloor\:\mathrm{for}\:\left({a},\:{b}\right)\in\mathbb{Z} \\$$$${then}\:{my}\:{answer}\:{is}\:{true} \\$$$$\mathrm{I}\:\mathrm{will}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{post}! \\$$$$\\$$$$\mathrm{if}\:\:{a},\:{b}\in\mathbb{Z},\:\lfloor{a}\rfloor={a},\:\lfloor{b}\rfloor={b} \\$$
Answered by Yozzi last updated on 24/Nov/15
$${Let}\:{y}=\lfloor\mathrm{1}−{x}\rfloor.\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor \\$$$${For}\:\mathrm{0}\leqslant{x}<\mathrm{1},\:\mathrm{0}<\mathrm{1}−{x}\leqslant\mathrm{1}\Rightarrow\mathrm{1}−{x}\:{is}\:{a}\: \\$$$${fraction}\:{or}\:\mathrm{1}\Rightarrow\lfloor\mathrm{1}−{x}\rfloor=\mathrm{0}\:{or}\:\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1} \\$$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor\:{is}\:{undefined}\:\left({cannot}\:{be}\:{sure}\:{to}\:{choose}\:\mathrm{1}\:{or}\:\mathrm{0}\right) \\$$$${For}\:−\mathrm{1}<{x}\leqslant\mathrm{0},\:\mathrm{1}\leqslant\mathrm{1}−{x}<\mathrm{2}\Rightarrow\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1}. \\$$$$\therefore\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{y}\neq\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}\:{since}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}\:{does}\:{not}\:{exist}. \\$$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{y}\:{does}\:{not}\:{exist}.\:{Graphically}, \\$$$${there}\:{is}\:{a}\:{jump}\:{discontinuity}\:{in} \\$$$${y}=\lfloor\mathrm{1}−{x}\rfloor\:{at}\:{x}=\mathrm{0}.\: \\$$$$\\$$