# sequence has a limit

#### Vali

##### Member
I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
I always get stuck at this kind of exercises.How to approach an exercise like this?

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#### Vali

##### Member
Sorry for my bad English, if that's what you mean..

#### Wilmer

##### In Memoriam
Sorry for my bad English, if that's what you mean..
No. Just a joke...
"walking on tip-toe" means "very carefully"!

#### Olinguito

##### Well-known member
I have the sequence from the picture and I have to demonstrate that this sequence has a limit.
I always get stuck at this kind of exercises.How to approach an exercise like this?
Hi Vali .

Unless I’ve missed something, I make it that the sequence does not converge.

Observe that
$$x_n^2=\left(x_{n-1}+\frac1{x_{n-1}}\right)^2=x_{n-1}^2+2+\frac1{x_{n-1}^2}>x_{n-1}^2+2.$$

Thus:
$$x_n^2>x_{n-1}^2+2>x_{n-2}^2+4>\cdots>x_0^2+2n>2n.$$

Thus $x_n>\sqrt{2n}\to\infty$ as $n\to\infty$ $\implies$ $x_n$ also $\to\infty$ as $n\to\infty$.

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#### Vali

##### Member
I tried to calculate x_(n+1) - x_(n) which is 1/x_(n) which is positive so x_(n) increases.Then, I assumed that x_(n)>0 and demonstrate that x_(n+1) > 0 but x_(n+1)=x_(n)+1/x_(n) so x_(n+1)>0 and now I don't know how to continue.

#### DavidCampen

##### Member
Olinguito gave you the complete answer.

As an aside, not for pre-calculus, I notice that the difference between any 2 consecutive terms is a convergent of [a; a, a. ...] where a = x_0 . Is this correct?

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#### Country Boy

##### Well-known member
MHB Math Helper
I would start by assuming that a limit exists and determining what that limit must be. Calling the limit "X" and taking the limit on both sides of $$\displaystyle x_{n+1}= x+\frac{1}{x_n}$$ we get $$\displaystyle A= A+ \frac{1}{A}$$. That reduces to $$\displaystyle \frac{1}{A}= 0$$ which is not true for any A! As Olinguito said, this sequence does not converge.

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