- #1

kindlychung

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## Homework Statement

1. Prove that the sequenced defined by [tex]x_{1}=3 [/tex] and [tex] x_{n+1} = \frac{1}{4-x_{n}}[/tex] converges.

2. Now that we know [tex] \lim x_{n}[/tex] exists, explain why [tex] \lim x_{n+1}[/tex] must exist and equal the same value.

3. Take the limit of each side of the recursive equation in part 1 of this exercise to explicitly compute [tex] \lim

## Homework Equations

[tex]x_{1}=3 [/tex] and [tex] x_{n+1} = \frac{1}{4-x_{n}}[/tex]

## The Attempt at a Solution

1. If we try to crunch each term of [tex] x_{n}[/tex] starting from [tex] x_{1}[/tex], we can easily see that [tex] x_{n}[/tex] is monotone and bounded, hence convergent. The the recursive form of fraction is rather annoying, makes it hard to prove.

2. Let [tex] \lim x_{n} = x[/tex], then there is an N, such that whenever [tex] n \ge N[/tex], we have [tex] |x_{n} - x | < \epsilon [/tex] for any positive [tex] \epsilon[/tex], but n+1 > n > N, hence [tex] |x_{n+1} - x | < \epsilon [/tex], as desired.

3. Let [tex] \lim x_{n} = \lim x_{n+1} = x[/tex], take limits of both sides, we have [tex] x = \frac{1}{4-x}[/tex], solve the equation we get the limit x.