Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 2624 by prakash jain last updated on 23/Nov/15

Old question related to greatest int function. lim_(x→0) ⌊1+x⌋=1 ⌊1⌋=1 lim_(x→0) ⌊1−x⌋=?

$$\mathrm{Old}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{greatest}\:\mathrm{int}\:\mathrm{function}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}+{x}\rfloor=\mathrm{1} \\ $$$$\lfloor\mathrm{1}\rfloor=\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=? \\ $$

Answered by Filup last updated on 24/Nov/15

if ⌊a±b⌋=⌊a⌋±⌊b⌋: for (a, b)∈Z lim_(x→0) ⌊1+x]=lim_(x→0) ⌊1⌋+lim_(x→0) ⌊x⌋ =1+lim_(x→0) ⌊x⌋=1+⌊0⌋=1 Similarly lim_(x→0) ⌊1−x⌋=lim_(x→0) ⌊1⌋−lim_(x→0) ⌊x⌋ =1−⌊0⌋=1

$$\mathrm{if}\:\lfloor{a}\pm{b}\rfloor=\lfloor{a}\rfloor\pm\lfloor{b}\rfloor:\:\:\:\:\:\mathrm{for}\:\:\left({a},\:{b}\right)\in\mathbb{Z} \\ $$$$ \\ $$$$\left.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}+{x}\right]=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}\rfloor+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor \\ $$$$=\mathrm{1}+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor=\mathrm{1}+\lfloor\mathrm{0}\rfloor=\mathrm{1} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Similarly}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}−{x}\rfloor=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor\mathrm{1}\rfloor−\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{x}\rfloor \\ $$$$=\mathrm{1}−\lfloor\mathrm{0}\rfloor=\mathrm{1} \\ $$

Commented by Yozzi last updated on 24/Nov/15

Let a=1.99,b=1.01 ⇒⌊1.99+1.01⌋=⌊3⌋=3≠⌊1.99⌋+⌊1.01⌋=1+1=2

$${Let}\:{a}=\mathrm{1}.\mathrm{99},{b}=\mathrm{1}.\mathrm{01} \\ $$$$\Rightarrow\lfloor\mathrm{1}.\mathrm{99}+\mathrm{1}.\mathrm{01}\rfloor=\lfloor\mathrm{3}\rfloor=\mathrm{3}\neq\lfloor\mathrm{1}.\mathrm{99}\rfloor+\lfloor\mathrm{1}.\mathrm{01}\rfloor=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$ \\ $$

Commented by Filup last updated on 24/Nov/15

Thanks for the proof! Back to the drawing board!

$${T}\mathrm{hanks}\:\mathrm{for}\:\mathrm{the}\:\mathrm{proof}! \\ $$$${B}\mathrm{ack}\:\mathrm{to}\:\mathrm{the}\:\mathrm{drawing}\:\mathrm{board}! \\ $$

Commented by Filup last updated on 24/Nov/15

if ⌊a±b⌋=⌊a⌋±⌊b⌋ for (a, b)∈Z then my answer is true I will edit my post! if a, b∈Z, ⌊a⌋=a, ⌊b⌋=b

$${if}\:\:\lfloor{a}\pm{b}\rfloor=\lfloor{a}\rfloor\pm\lfloor{b}\rfloor\:\mathrm{for}\:\left({a},\:{b}\right)\in\mathbb{Z} \\ $$$${then}\:{my}\:{answer}\:{is}\:{true} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{post}! \\ $$$$ \\ $$$$\mathrm{if}\:\:{a},\:{b}\in\mathbb{Z},\:\lfloor{a}\rfloor={a},\:\lfloor{b}\rfloor={b} \\ $$

Answered by Yozzi last updated on 24/Nov/15

Let y=⌊1−x⌋. lim_(x→0^+ ) y=lim_(x→0^+ ) ⌊1−x⌋ For 0≤x<1, 0<1−x≤1⇒1−x is a fraction or 1⇒⌊1−x⌋=0 or ⌊1−x⌋=1 ⇒lim_(x→0^+ ) ⌊1−x⌋ is undefined (cannot be sure to choose 1 or 0) For −1<x≤0, 1≤1−x<2⇒⌊1−x⌋=1. ∴lim_(x→0^− ) ⌊1−x⌋=1. lim_(x→0^− ) y≠lim_(x→0^+ ) y since lim_(x→0^+ ) y does not exist. ⇒lim_(x→0) y does not exist. Graphically, there is a jump discontinuity in y=⌊1−x⌋ at x=0.

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com