I've got two calculus proofs that I can't seem to get! I was wondering if you guys could help me out a bit...(adsbygoogle = window.adsbygoogle || []).push({});

1.

The problem statement, all variables and given/known data

Suppose [tex]x_{n}[/tex] is the sequence defined recursively by

[tex]x_{1}=0[/tex] and [tex]x_{n+1}=\sqrt{5 + 2x_{n}}[/tex] for [tex]n=0, 1, 2, 3, ....[/tex]

Prove that [tex]x_{n}[/tex] converges and find its limit

The attempt at a solution

So far, by taking the limit of both sides "[tex] l [/tex]", I've found the limit

[tex]l=\sqrt{5 + 2l}[/tex]

[tex]l=\frac{-2 +\sqrt{24}}{2}[/tex]

but I still have to prove that the sequence converges. I've tried induction hypothesis using

[tex] x_{n} \leq x_{n+1} \leq \frac{-2 +\sqrt{24}}{2}[/tex]

and then I attempt to modify [tex]x_{n}[/tex] and [tex]x_{n+1}[/tex] such that the equation becomes (in the end):

[tex] x_{n+1} \leq x_{n+2} \leq \frac{-2 +\sqrt{24}}{2}[/tex]

but I can never get it to work out!

2.

The problem statement, all variables and given/known data

The second proof I'm having trouble with is basically one of the same type...

Let [tex]x_{n}[/tex] be the sequence of real numbers defined recursively by

[tex]x_{0} = 0 [/tex] and [tex] x_{n+1} = \ln (2 + x_{n}) [/tex] for [tex] n=0, 1, 2, 3,..... [/tex]

Show that [tex] x_{n} \leq x_{n+1} \leq 2 [/tex]

The attempt at a solution

Again, I try to modify it such that i get

[tex] \ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq 2 [/tex]

but I keep ending up with

[tex] \ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq \ln (2 + 2) [/tex]

Is there a trick to calculus proofs?

Thanks!

Steve

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# Homework Help: Hard calculus proofs!

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