Find the limit of the following sequence:
2 , 2 2 , 2 2 2 , …
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Well you first have to show the limit exists.
Log in to reply
Nice solution but how to show the limit exits , Please help.
Log in to reply
the limit exists at 2.
Log in to reply
@Lipsa Kar – You'll need to prove that though.
Log in to reply
@Daniel Liu – correct me if i'm wrong, we need to prove (2x)^1/2 limit exists at 2 or not. {2(2+h)}^1/2= root 2 x root of (2+h)= 2= rhs similarly, lhs=2 and putting value of 2 in it, we get 2. lhs=rhs=f(2). hence, limit exists at 2.
Log in to reply
@Lipsa Kar – Sorry, but I don't really understand your explanation. Also, what do you mean by "limit exists at 2"? Do you mean "the limit exists, and it's value is 2"?
Log in to reply
@Daniel Liu – Can you please give your explanation and method of proving that limit exist.
@Daniel Liu – what i meant was the limit of that function exists at x=2. where the function is f(x). how do you show the limit exists or not at that point? we consider the left hand limit and the right hand limit. f(x) is there, then to know the left hand limit, which we show as lim (x tends to 2 minus but not exactly 2, just before 2) which we take as f(2-h) where h tends to 0. we then solve that, and after solving everything, i find that, f(2)= f(2-h)=f(2+h) where h tends to 0. i have just applied the old common method. what i meant by limit exists at 2 is that the limit exists at x=2 where the function is f(x). i know my explanation is not that great but i hope i'm clear now. :)
Log in to reply
@Lipsa Kar – Ah no, I'm not exactly sure what you are doing and what the function you are talking about is, but I'm fairly sure we are talking about different things.
What I said required proof was this: Given the sequence a 0 = 1 and a n = 2 a n − 1 , prove that n → ∞ lim a n exists.
Log in to reply
@Daniel Liu – Can you please go with this limiting from beginning step by step until end , I really love to add that in my Brain...
I want to become Math Champ ! :D
Log in to reply
@Syed Baqir – See Otto Bretscher's explanation below.
Log in to reply
@Daniel Liu – Thanks , If you have some other explanation please dont hesitate to share with us.
Cheers, :D
@Daniel Liu – so sorry, my function wasn't correct! yours is the correct solution! thanks for helping out :)
A^2=2A A=0,2 both. ......... So why is "0" not an answer??
Log in to reply
2^x is continuous and as x -> infinity , limit approaches 2 not 0.
Just working out the first few terms of the series, the sequence soon exceeds 2 and each successive term gets higher than the previous. The solution offered is asking us to potentially divide infinity by infinity. The series diverges.
The n th term of the sequence is 2 2 1 + 4 1 + . . . + 2 n 1 , and the limit is 2 1 = 2
Nice!
Sir, I am little puzzled because I am wondering how : \sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2.... } } } } } \Longrightarrow { 2 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\quad ...\quad +\quad \frac { 1 }{ { 2 }^{ n } } }
Log in to reply
Let's look at the third term of the sequence, 2 2 2 , for example. Repeatedly using the rule a b = a b , we find 2 2 2 = 2 4 2 8 2 = 2 2 1 + 4 1 + 8 1 . To formally prove the analogous formula 2 2 1 + 4 1 + . . . + 2 n 1 for the n th term, use induction. Note that lim n → ∞ ( 2 1 + 4 1 + . . . + 2 n 1 ) = 1 , a geometric series. Now lim n → ∞ 2 2 1 + 4 1 + . . . + 2 n 1 = 2 1 = 2 since 2 x is continuous at x = 1 .
This is the best solution; it takes care of the limit and its existence at the same time.
Let the sequence be written as a 1 , a 2 , a 3 , … . The first term is a 1 = 2 2 1 and each subsequence term is a n + 1 = 2 a n . Let a n = 2 b n . Using this, we have 2 b n + 1 ⟹ b n + 1 = a n + 1 = 2 × a n = 2 × 2 b n = 2 2 b n + 1 = 2 b n + 1 .
Now let c n = 1 − b n . From the above, we get 1 − c n + 1 ⟹ c n + 1 = 2 1 − c n + 1 = 1 − 2 c n = 2 c n
Since a 1 = 2 and hence b 1 = 2 1 , then c 1 = 2 1 and lim n → ∞ c n = 0 . Therefore lim n → ∞ b n = lim n → ∞ 1 − c n = lim n → ∞ 1 − 0 = 1 . Finally, this results in n → ∞ lim a n = n → ∞ lim 2 b n = 2 lim n → ∞ b n = 2 1 = 2
Nice solution but last paragraph is a little vague for me or maybe for some other ppl could you please elaborate more ,
Espiecially , lim Bn = lim (1-Cn) => lim ( 1 - 0 ) = 1
And also , lim An = 2^(Bn)
Thank you very Much
"Upvoted"
Log in to reply
Well from before, I defined a n = 2 b n and b n = 1 − c n so I substituted them in.
The limit pf this sequence converges to 2.
By squaring both sides of equation x = 2 2 2 … and then dividing by two we get: x 2 / 2 = 2 2 2 … = x . That is x 2 = 2 x and therefore x = 2 .
x can be 0 too!
Log in to reply
under the root, there is a non-zero value always. and nothing is added or subtracted, things here are just multiplied. we can only get the answer 0 if there was a term 0 in the series. so, we are taking 2 as the answer.
Log in to reply
Can you then explain why 0 came mathematically?
Log in to reply
@Kishore S. Shenoy – we are considering the value of that series to be x(say) and then we see that under the root, the same series is being multiplied with 2. mathematically, since x is on both rhs and lhs and it is multiplied, x has a value 0 (x=(2x)^1/2) and 0 satisfies the equation mathematically. but here, in actual, we know that 0 can never be the value of x since all it has is non-zero numbers multiplied with each other. another random example, root of 4 is always 2 but 4 can be written as (-2)(-2). only positive roots are being considered. mathematically true, but not actual
Log in to reply
@Lipsa Kar – Oh, that multiplication! Thanks!
Problem Loading...
Note Loading...
Set Loading...
The sequence of 2 goes forever. So the limits of this is:
A = 2 2 2 2 … ⇒ A 2 = 2 × 2 2 2 2 … A 2 = 2 × A ⇒ A A 2 = 2 ⇒ A = 2