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

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

\underset{x \to 0}{l i m} \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x \dots \dots .}}}}

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 .

Commented by DavidClassic last updated on 16/Jun/20

l e t u = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots .}}}}

u^{2} = x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + . .}}}}

u^{2} = x + u

u^{2} - u - x = 0

u = \frac{1 \pm \sqrt{1 - 4 x^{2}}}{2}

_{x \to 0}^{l i m} u = \frac{1 \pm \sqrt{1 - 0}}{2} = \frac{1 \pm 1}{2} = 1 o r 0

b y : O g u n s a n w o D a v i d O . C o u r t e s y : D a v e C l a s s i c s

Commented by Aziztisffola last updated on 16/Jun/20

x and u (x) are positive then u (x) = \frac{1 + \sqrt{1 + 4 x}}{2}

the limit is unique is equal 1

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

T i n k u t a r a, t h e d e v e l o p e r

S i r, I^{'} v e a p r o b l e m . I n e d i t i n g

m o d e i f t h e s c r e e n i s t u r n e d

i n t o l a n d s c a p e u n i n t e n t i o n a l l y

a n d I t u r n b a c k i n t o p o r t r a i t

m o d e t h e d o w n h a l f s c r e e n i s

c l e a r e d! I t c o n t i n u e s i n v a r i o u s

v h e r s i o n s e v e n i n 2.084 . T h a n k s .

S e e s c r e e n s h o t b e l o w .

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 .

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

Y e s s i r e v e r y t i m e .

P h o n e m a k e : V i v o

M o d e l : V i v o S 1

a n d r o i d v e r s i o n : 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

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

i h a v e t h e s a m e p r o b l e m . b u t f o r m e

t h e e d i t o r i n l a n d s c a p e m o d u s i s n o t

a r e a l h e l p e r .

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 .

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 .

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 .

Answered by Aziztisffola last updated on 16/Jun/20

f (x) = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{\dots . .}}}}

\Leftrightarrow f^{2} (x) = x + f (x) \Leftrightarrow f^{2} (x) - f (x) - x = 0

△ = 1 + 4 x \Rightarrow f (x) = \frac{1 + \sqrt{1 + 4 x}}{2} (x > 0 and f (x) > 0)

\underset{x \to 0}{\lim f} (x) = \frac{1 + \sqrt{1}}{2} = 1

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

\underset{x \to 0^{+}}{\lim f} (x) = 1 \neq f (0) = 0

Answered by MAB last updated on 16/Jun/20

v e r y i n f o r m a l r e s o l u t i o n

l e t u = \sqrt{x + \sqrt{x + \sqrt{x \dots}}}

w e h a v e

u = \sqrt{x + u}

i n o t h e r w o r d s

u^{2} - u - x = 0

t h e n

u = \frac{1 \mp \sqrt{1 + 4 x}}{2}

a s x \to 0 w e g e t

u = 0 o r u = 1

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

let y (x) = \sqrt{x + \sqrt{x + \sqrt{x} + \dots}}

\Rightarrow y^{2} (x) = x + y (x) with x ⩾ 0 \Rightarrow y^{2} - y - x = 0

Δ = 1 + 4 x > 0 \Rightarrow y_{1} = \frac{1 + \sqrt{1 + 4 x}}{2} and y_{2} = \frac{1 - \sqrt{1 + 4 x}}{2} (y_{2} < 0 to eliminate) \Rightarrow

y (x) = \frac{1 + \sqrt{4 x + 1}}{2} \Rightarrow \lim_{x \to 0} y (x) = 1