lim-x-0-x-x-x-x-

Question Number 98858 by M±th+et+s last updated on 16/Jun/20

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…….}}}} \\ $$$$ \\ $$

Commented by Tinku Tara last updated on 17/Jun/20

Hi David Can you update app to latesy version. Some time un between problem related to reducing font size in successive edit was fixed.

$$\mathrm{Hi}\:\mathrm{David} \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{update}\:\mathrm{app}\:\mathrm{to}\:\mathrm{latesy}\:\mathrm{version}. \\ $$$$\mathrm{Some}\:\mathrm{time}\:\mathrm{un}\:\mathrm{between}\:\mathrm{problem} \\ $$$$\mathrm{related}\:\mathrm{to}\:\mathrm{reducing}\:\mathrm{font}\:\mathrm{size} \\ $$$$\mathrm{in}\:\mathrm{successive}\:\mathrm{edit}\:\mathrm{was}\:\mathrm{fixed}. \\ $$

Commented by DavidClassic last updated on 16/Jun/20

let u=(√(x+(√(x+(√(x+(√(x+....)))))))) u^2 =x+(√(x+(√(x+(√(x+(√(x+..)))))))) u^2 =x+u u^2 −u−x=0 u=((1±(√(1−4x^2 )))/2) _(x→0)^(lim) u=((1±(√(1−0)))/2)=((1±1)/2)=1 or 0 by: Ogunsanwo David O. Courtesy: Dave Classics

$${let}\:{u}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+….}}}} \\ $$$${u}^{\mathrm{2}} ={x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+..}}}} \\ $$$${u}^{\mathrm{2}} ={x}+{u} \\ $$$${u}^{\mathrm{2}} −{u}−{x}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:_{{x}\rightarrow\mathrm{0}} ^{{lim}} {u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{0}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:{or}\:\mathrm{0} \\ $$$${by}:\:\mathrm{O}{gunsanwo}\:{David}\:{O}.\:\:\:{Courtesy}:\:\mathrm{D}{ave}\:{Classics} \\ $$

Commented by Aziztisffola last updated on 16/Jun/20

x and u(x) are positive then u(x)=((1+(√(1+4x)))/2) the limit is unique is equal 1

$$\:\mathrm{x}\:\mathrm{and}\:\mathrm{u}\left(\mathrm{x}\right)\:\mathrm{are}\:\mathrm{positive}\:\mathrm{then}\:\mathrm{u}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}} \\ $$$$\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{unique}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{1} \\ $$

Commented by Rasheed.Sindhi last updated on 17/Jun/20

Tinkutara,the developer Sir,I′ve a problem.In editing mode if the screen is turned into landscape unintentionally and I turn back into portrait mode the down half screen is cleared! It continues in various vhersions even in 2.084.Thanks. See screenshot below.

$$\mathcal{T}{inkutara},{the}\:{developer} \\ $$$${Sir},{I}'{ve}\:{a}\:{problem}.{In}\:{editing} \\ $$$${mode}\:{if}\:{the}\:{screen}\:{is}\:{turned} \\ $$$${into}\:{landscape}\:{unintentionally} \\ $$$${and}\:{I}\:{turn}\:{back}\:{into}\:{portrait} \\ $$$${mode}\:{the}\:{down}\:{half}\:{screen}\:{is} \\ $$$${cleared}!\:{It}\:{continues}\:{in}\:{various} \\ $$$${vhersions}\:{even}\:{in}\:\mathrm{2}.\mathrm{084}.{Thanks}. \\ $$$${See}\:{screenshot}\:{below}. \\ $$

Commented by Rasheed.Sindhi last updated on 17/Jun/20

Commented by Tinku Tara last updated on 17/Jun/20

Does it happen every time? Please email phone make and model, android versiin.

$$\mathrm{Does}\:\mathrm{it}\:\mathrm{happen}\:\mathrm{every}\:\mathrm{time}? \\ $$$$\mathrm{Please}\:\mathrm{email}\:\mathrm{phone}\:\mathrm{make}\:\mathrm{and}\:\mathrm{model}, \\ $$$$\mathrm{android}\:\mathrm{versiin}. \\ $$

Commented by Rasheed.Sindhi last updated on 17/Jun/20

Yes sir every time. Phone make: Vivo Model: Vivo S1 android version:9

$${Yes}\:{sir}\:{every}\:{time}. \\ $$$${Phone}\:{make}:\:{Vivo} \\ $$$${Model}:\:{Vivo}\:{S}\mathrm{1} \\ $$$${android}\:{version}:\mathrm{9} \\ $$

Commented by MJS last updated on 17/Jun/20

same problem here. but the lines are not lost, it′s just impossible to scroll there. if you save the post they are there again

$$\mathrm{same}\:\mathrm{problem}\:\mathrm{here}.\:\mathrm{but}\:\mathrm{the}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{lost},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{scroll}\:\mathrm{there}.\:\mathrm{if} \\ $$$$\mathrm{you}\:\mathrm{save}\:\mathrm{the}\:\mathrm{post}\:\mathrm{they}\:\mathrm{are}\:\mathrm{there}\:\mathrm{again} \\ $$

Commented by mr W last updated on 17/Jun/20

$${i}\:{have}\:{the}\:{same}\:{problem}.\:{but}\:{for}\:{me} \\ $$$${the}\:{editor}\:{in}\:{landscape}\:{modus}\:{is}\:{not} \\ $$$${a}\:{real}\:{helper}. \\ $$

Commented by Tinku Tara last updated on 17/Jun/20

Ok. So the problem occurs when you are rotating phone in middle of editing. Mostly handling is missing for these cases. Will fix.

$$\mathrm{Ok}.\:\mathrm{So}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{occurs} \\ $$$$\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{rotating}\:\mathrm{phone}\:\mathrm{in} \\ $$$$\mathrm{middle}\:\mathrm{of}\:\mathrm{editing}.\:\mathrm{Mostly}\:\mathrm{handling} \\ $$$$\mathrm{is}\:\mathrm{missing}\:\mathrm{for}\:\mathrm{these}\:\mathrm{cases}. \\ $$$$\mathrm{Will}\:\mathrm{fix}. \\ $$

Commented by Tinku Tara last updated on 17/Jun/20

landscape mode request came from some school where rotation of device in portrait mode was not possible. It is not really useful for typing on phones.

$$\mathrm{landscape}\:\mathrm{mode}\:\mathrm{request}\:\mathrm{came}\:\mathrm{from} \\ $$$$\mathrm{some}\:\mathrm{school}\:\mathrm{where}\:\mathrm{rotation}\:\mathrm{of} \\ $$$$\mathrm{device}\:\mathrm{in}\:\mathrm{portrait}\:\mathrm{mode}\:\mathrm{was}\:\mathrm{not} \\ $$$$\mathrm{possible}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{not}\:\mathrm{really}\:\mathrm{useful}\:\mathrm{for}\:\mathrm{typing} \\ $$$$\mathrm{on}\:\mathrm{phones}. \\ $$

Commented by Tinku Tara last updated on 26/Jun/20

This issue has been addressed in latest version 2.085. Device rotation will not rotate editor screen. Let us know if the problem still exists.

$$\mathrm{This}\:\mathrm{issue}\:\mathrm{has}\:\mathrm{been}\:\mathrm{addressed}\:\mathrm{in} \\ $$$$\mathrm{latest}\:\mathrm{version}\:\mathrm{2}.\mathrm{085}. \\ $$$$\mathrm{Device}\:\mathrm{rotation}\:\mathrm{will}\:\mathrm{not} \\ $$$$\mathrm{rotate}\:\mathrm{editor}\:\mathrm{screen}.\: \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{know}\:\mathrm{if}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{still} \\ $$$$\mathrm{exists}. \\ $$

Answered by Aziztisffola last updated on 16/Jun/20

f(x)=(√(x+(√(x+(√(x+(√(.....)))))))) ⇔f^2 (x)=x+f(x)⇔f^2 (x)−f(x)−x=0 △=1+4x ⇒f(x)=((1+(√(1+4x)))/2) (x>0 and f(x)>0) lim_(x→0) f(x)=((1+(√1))/2)=1

$$\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{…..}}}} \\ $$$$\:\Leftrightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}+\mathrm{f}\left(\mathrm{x}\right)\Leftrightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{f}\left(\mathrm{x}\right)−\mathrm{x}=\mathrm{0} \\ $$$$\bigtriangleup=\mathrm{1}+\mathrm{4x}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\left(\mathrm{x}>\mathrm{0}\:\mathrm{and}\:\mathrm{f}\left(\mathrm{x}\right)>\mathrm{0}\right) \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}f}\left(\mathrm{x}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}}}{\mathrm{2}}=\mathrm{1} \\ $$

Commented by mr W last updated on 16/Jun/20

$$\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}f}\left(\mathrm{x}\right)=\mathrm{1}\:\neq\:\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$

Answered by MAB last updated on 16/Jun/20

very informal resolution let u=(√(x+(√(x+(√(x...)))))) we have u=(√(x+u)) in other words u^2 −u−x=0 then u=((1∓(√(1+4x)))/2) as x→0 we get u=0 or u=1

$${very}\:{informal}\:{resolution} \\ $$$${let}\:{u}=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}} \\ $$$${we}\:{have} \\ $$$${u}=\sqrt{{x}+{u}} \\ $$$${in}\:{other}\:{words} \\ $$$${u}^{\mathrm{2}} −{u}−{x}=\mathrm{0} \\ $$$${then} \\ $$$${u}=\frac{\mathrm{1}\mp\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}} \\ $$$${as}\:{x}\rightarrow\mathrm{0}\:{we}\:{get} \\ $$$${u}=\mathrm{0}\:{or}\:{u}=\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 16/Jun/20

let y(x) =(√(x+(√(x+(√x)+...)))) ⇒y^2 (x) =x+y(x) with x≥0 ⇒y^2 −y−x =0 Δ =1+4x >0 ⇒y_1 =((1+(√(1+4x)))/2) and y_2 =((1−(√(1+4x)))/2)(y_2 <0 to eliminate) ⇒ y(x) =((1+(√(4x+1)))/2) ⇒lim_(x→0) y(x) =1

$$\mathrm{let}\:\mathrm{y}\left(\mathrm{x}\right)\:=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}+…}} \\ $$$$\Rightarrow\mathrm{y}^{\mathrm{2}} \left(\mathrm{x}\right)\:=\mathrm{x}+\mathrm{y}\left(\mathrm{x}\right)\:\:\mathrm{with}\:\mathrm{x}\geqslant\mathrm{0}\:\Rightarrow\mathrm{y}^{\mathrm{2}} −\mathrm{y}−\mathrm{x}\:=\mathrm{0} \\ $$$$\Delta\:=\mathrm{1}+\mathrm{4x}\:>\mathrm{0}\:\Rightarrow\mathrm{y}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{y}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{1}+\mathrm{4x}}}{\mathrm{2}}\left(\mathrm{y}_{\mathrm{2}} <\mathrm{0}\:\mathrm{to}\:\mathrm{eliminate}\right)\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}+\sqrt{\mathrm{4x}+\mathrm{1}}}{\mathrm{2}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\mathrm{y}\left(\mathrm{x}\right)\:=\mathrm{1} \\ $$

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