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What is the limit of $U_{n+1} = \dfrac{2U_n + 3}{U_n + 2}$ and $U_0 = 1$?

I need the detail, and another way than using the solution of $f(x)=x$, as $f(x) = \frac{2x+3}{x+2}$ because I can't show that $f(I) \subseteq I$ as $I = ]-\infty; -2[~\cup~ ]-2; +\infty[$.

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    You can write $U_{n+1}$ in LaTeX as `$U_{n+1}$`.2012-11-02
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    yeah I mistype it2012-11-02
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    What does your $I$ have to do with the problem? The starting point is $1\notin I$ (and as a matter of fact all later points are $\notin I$ as well).2012-11-02

6 Answers 6