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

$${Let}\:{y}=\lfloor\mathrm{1}−{x}\rfloor.\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor \\ $$$${For}\:\mathrm{0}\leqslant{x}<\mathrm{1},\:\mathrm{0}<\mathrm{1}−{x}\leqslant\mathrm{1}\Rightarrow\mathrm{1}−{x}\:{is}\:{a}\: \\ $$$${fraction}\:{or}\:\mathrm{1}\Rightarrow\lfloor\mathrm{1}−{x}\rfloor=\mathrm{0}\:{or}\:\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor\:{is}\:{undefined}\:\left({cannot}\:{be}\:{sure}\:{to}\:{choose}\:\mathrm{1}\:{or}\:\mathrm{0}\right) \\ $$$${For}\:−\mathrm{1}<{x}\leqslant\mathrm{0},\:\mathrm{1}\leqslant\mathrm{1}−{x}<\mathrm{2}\Rightarrow\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1}. \\ $$$$\therefore\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=\mathrm{1}.\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{y}\neq\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}\:{since}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{y}\:{does}\:{not}\:{exist}. \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{y}\:{does}\:{not}\:{exist}.\:{Graphically}, \\ $$$${there}\:{is}\:{a}\:{jump}\:{discontinuity}\:{in} \\ $$$${y}=\lfloor\mathrm{1}−{x}\rfloor\:{at}\:{x}=\mathrm{0}.\: \\ $$$$ \\ $$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com