sqrt(0+sqrt(0+sqrt(0+...))) = ?

sqrt(0+sqrt(0+sqrt(0+...)))=0 but what will be the limit as x goes to 0+ of sqrt(x+sqrt(x+sqrt(x+...)))? This is a very intriguing calculus limit. So how do we evaluate this limit with infinitely many nested square roots? This limit is actually a limit indeterminate form. The key here is that the limit of a sequence of continuous functions might not be continuous.

0:00 Intro
7:12 Graphs of sqrt(x), sqrt(x+sqrt(x)), ...

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[averageanon]

you can gradually see his beard evolving to archimedes

[sudoheckbegula]

Finally, the man on HIS actual channel!!

[ЕвгенийСизов-хг]

I like how the “1” in the answer differs from all the others, like it’s the most special one

[Fysiker]

Fun argument, I wasn't sure how to set it up rigorously. I guessed 1 by thinking that for 0<x<1 that √x > x, and that nesting √x in more square roots would make it increase all the way to 1 because you keep square rooting a number less than 1. Even if adding the little x every nesting managed to increase the value to above 1, square rooting a number greater than 1 will decrease its value until it reaches 1.

[blackpenredpen]

Remember when you square both sides, you might get extraneous solutions.

[leumasarc]

Blackpenredpen: As always, pause the video and try it first!

Me: No, I don't think I will...

[sajgol]

Hey one thing to confess.
Did you know that you are awesome?
Even paying attention to the tiny details. In our schools they would have just substituted zero to the x in quadratic answers and tada that's the answer

[nashaut]

Probably the coolest demonstration I've seen! Absolutely Fabulous!

[jimschneider]

I had to run a few numerical experiments to convince myself I hadn't made a big dumb error in the algebra. This is one of those nice problems that makes you think, and then makes you check your thinking.

[stevemonkey]

Good to see a video where you are speaking slowly. 😁👍

[ciaranmanca]

I'm doing a calculus course now. This man makes it look so easy

[devanshsharma]

oh man that sequence thing. never thought as that. caught me totally off gaurd. such a good video. thnx bprp

[jarikosonen]

It looked at some point that L=(1-sqrt(1+4*x))/2 is the solution needed at x=0 [ giving L=0 ].

[brumarul]

Exactly what I needed, I had a limit of (an)n>= 1, where a0 = sqrt(3) and an+1 = sqrt(3+2an), this video made me solve it, I struggled with it for a while, but you made it clear.

[cmilkau]

At 5:55, you can only drop the minus if X is strictly positive. if x is 0, as in Q1, you get two solutions. (0, 1). This makes sense because 0 and 1 both are fixed points of the square root. However, 0 is unstable and 1 is stable, so the Q2 limit approaches only 1.

[e-learningtutor]

Its insane..enjoyed watching
And waiting for some more like these :)

I done seened it, but I still can't believe it...

(surely it's -1/12). 😁

[NubPaws]

Interesting video! That's a nice way of looking at a problem. I'll keep that as a mental note. Thanks for another interesting video!

[SeekingTheLoveThatGodMeans]

As long as x is above 0 at all, the sequence of square roots will grow towards 1 as the infinitely iterative square rooting process amplifies that infinitesimal number way out at infinity to 1-. This behavior can be seen by carrying it out on a calculator with the smallest positive x that it can represent, and pretty soon the original x is virtually immaterial to the result of the sequence (with floating point, adding it in just underflows anyhow).

[Yggmm-qh]

Wow! Thank you for solve what I wondered question. When I draw this two graph, I confused that they're not same. So I asked my friend and math teacher, but no one solved this question XD