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- Thread starter Vali
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...by walking on tip-toeHow to approach an exercise like this?

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No. Just a joke...Sorry for my bad English, if that's what you mean..

"walking on tip-toe" means "very carefully"!

- Apr 22, 2018

- 251

Hi Vali .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?

Unless I’ve missed something, I make it that the sequence does

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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- Apr 4, 2014

- 64

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?

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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- Jan 30, 2018

- 371

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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