Is this analysis problem or discrete math problem?

Suppose $x_{n} \in \mathbb{R}, x_{1} = 1, 2x_{n+1} = x_{n} + 3/x_{n}$. Then show that $\lim x_{n}$ exists and find its value.

So is this problem (real) analysis problem or a discrete math one?

Could you suggest me a good textbook that deals with this kind of problems?

As for the solution of this problem, I'd love to get a solution but if you can't be bothered then you don't have to.

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This is an analysis problem: basic calculus, in fact. –  DonAntonio Jan 19 '13 at 22:18
If you know about Newton's Method in Calculus, you could apply it to the equation $x^2-3=0$, and see what happens. –  Gerry Myerson Jan 19 '13 at 22:31
You could consider it a discrete math problem to prove that ineqalities hold for this sequence, e.g., that $x_n\geq 1$ for all $n$, and $x_{n+1}<x_n$ for all $n>1$. For that you don't really need real analysis, and you are working with rational numbers. But real analysis comes in where you want to prove that the limit exists (e.g., using the fact that the sequence is bounded and eventually monotone) and find its value. –  Jonas Meyer Jan 19 '13 at 22:36
This is a sweeping generalization, but limits are generally considered be a mathematical concept in the realm of calculus. –  Christopher A. Wong Jan 20 '13 at 0:06

First, we will show that $\{x_n\}$ is strictly decreasing and bounded (not counting the first term). To do this, we must show that $x_{n+1} = \frac{x_{n}}{2} + \frac{3}{2x_n} < x_n,$ which is equivalent to $x_n > \frac{3}{x_n},$ i.e. $x_n > \sqrt{3}.$ This last statement we can prove easily (for $n \ge 2$). By AM-GM, $x_{n+1} = \frac{x_n}{2} + \frac{3}{2x_n} \ge \sqrt{3},$ with equality iff $x_n = \sqrt{3}.$ Since $x_n \not= \sqrt{3}$ for $n=2,$ equality never occurs, i.e. $x_n > \sqrt{3}$ for all $n \ge 2,$ from which we have $\sqrt{3} < x_{n+1} < x_n,$ as noted above.
By the monotone convergence theorem, $\{x_n\}$ has a limit, say $x$. Then, by the given, $2x = x + \frac{3}{x},$ from which $x = \sqrt{3}.$