Question

limit sqrt(x+sqrt(x+sqrt(x)))/sqrt(x) x->infinity without hopital rule

Answer 1

do you understand this ?
\(\Large \dfrac{\sqrt{a+b}}{c} = \sqrt{\dfrac{a}{c^2}+\dfrac{b}{c^2}}\)
??

Answer 2

Let me see if I got this right...
\[\Large \lim_{x\rightarrow \infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x\sqrt x}\]

Answer 3

the denominator is only sqrt x

Answer 4

whoops... okay, fixing...\[\Large \lim_{x\rightarrow \infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt x}\]
Now then, @mononoor ready?

Answer 5

i Waiting

Answer 6

Right... then I suggest you put this property to good use...
\[\Large \sqrt{\frac{a}{b}}=\frac{\sqrt a}{\sqrt b}\]

Answer 7

waiting for what ? lol
sorry, this is not a free answering service.
its a learning site, and we can best learn when we interact with each other, agree ?

Answer 8

The gist of it is @mononoor I'm going to need you to work with me here ^_^

Answer 9

@hartnn I agree. but I understand this equation

Answer 10

so you understood the simplification that terenz and i suggested ?
can you try it for your question, your limit function ?

Answer 11

Or maybe just divide both numerator and denominator by \(\sqrt{x}\) first?

Answer 12

@RolyPoly I was going to say that, then I realised it's identical to just simplifying the fraction-radical... and it's less messy XD

Answer 13

Hehe! :D

Answer 14

\[\Large \lim_{x\rightarrow \infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt x}\]\[=\Large \lim_{x\rightarrow \infty}\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}\]Have fun with cancelling :D

Answer 15

Easy guys... I prefer not to begin before @mononoor does ;)

Answer 16

sqrt(x+sqrt(x+sqrt(x)))/sqrt(x)
sqrt(1+sqrt(x+sqrt(x))/x)

Answer 17

\[\lim_{x \rightarrow \infty} \frac{ \sqrt{x+\sqrt{x+\sqrt{x}}} }{ \sqrt{x} }=1\]
this is correct?

Answer 18

the final answer 1 ? how'd you get it?

Answer 19

i want to just verify that you've not guessed it, but actually solved it...

Answer 20

\[\lim_{x \rightarrow \infty} \sqrt{\frac{ x +\sqrt{x +\sqrt{x}} }{ x }}=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x +\sqrt{x}} }{ x }}=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x}\sqrt{x+1} }{ \sqrt{x}\sqrt{x} }}\]
\[=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x+1} }{ \sqrt{x} }} =\lim_{x \rightarrow \infty} \sqrt{1+\frac{\sqrt{x} \sqrt{1+1/\sqrt{x}} }{ \sqrt{x} }}=

Answer 21

you are on right direction, so i will believe you got it by actually solving :)
no need to show further steps if you don't want to...

Answer 22

\[=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x+1} }{ \sqrt{x} }} =\lim_{x \rightarrow \infty} \sqrt{1+\frac{\sqrt{x} \sqrt{1+1/\sqrt{x}} }{ \sqrt{x} }}=\]

Answer 23

\[\lim_{x \rightarrow \infty} \sqrt{1+ \sqrt{1+1/\sqrt{x}} }=\sqrt{1+\sqrt{1+0}}=\sqrt{1+1}=\sqrt{2}\]
oh... this slove to \[\sqrt{2}\]
sorry

Answer 24

actually there's an error in your working...

Answer 25

can you help me? Which stage؟

Answer 26

\(\Large \lim \limits_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x +\sqrt{x}} }{ x }}=\lim \limits_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x}\sqrt{\color{red}{\sqrt x}+1} }{ \sqrt{x}\sqrt{x} }}\)


\(\Large x+\sqrt x =\sqrt x (\sqrt x+1) \)

Answer 27

got this ?

Answer 28

oh yes.

Answer 29

any problems ? doubts anymore ?

Answer 30

no?

Answer 31

sorry. I busy at my office.
\[=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{x}\sqrt{\sqrt{x}+1} }{ \sqrt{x}\sqrt{x} }}=\lim_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{\sqrt{x}+1} }{ \sqrt{x} }}=?\]

Answer 32

if \[\sqrt{x}=a\] ==> \[1+\frac{ \sqrt{a+1} }{ a }=1+\frac{ 1 }{ \sqrt{a+1} }+\frac{ 1 }{ a \sqrt{a+1} }\] and \[x \rightarrow \infty \] then \[a \rightarrow \infty \]
\[\lim_{a \rightarrow \infty} \sqrt{1+\frac{ 1 }{ \sqrt{a+1} }+\frac{ 1 }{ a \sqrt{a+1} }}=\sqrt{1+0+0}=1\]
this is correct؟

Answer 33

yes! absolutely :)
good work!
and you could have done that without using 'a' to...

Answer 34

this is understanding for me. and thank from you.

Answer 35

\(\Large \lim \limits_{x \rightarrow \infty} \sqrt{1+\frac{ \sqrt{\color{red}{\sqrt x}+1} }{ \sqrt{x} }} =\lim \limits_{x \rightarrow \infty} \sqrt{1+\sqrt {\frac{{\color{red}{\sqrt x}+1} }{ {x} }} } =\lim \limits_{x \rightarrow \infty} \sqrt{1+\sqrt {\frac{{\color{red}{1}} }{ {\sqrt x} }+\frac{1}{x}} } \)
and since x -> infinity 1/sqrt x and 1/x = 0
just the same thing said differently.
oh, good,
welcome ^_^

Answer 36

since you are new here,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)
if you have any doubts browsing this site, you can ask me.
Have fun learning with Open Study! :)

Answer 37

very very happy to coming to this site. and my language not English. but I try to learn it.

Answer 38

we are also glad to have you here :)
you're doing good at learning english, keep it up :)