To prove existence of the limit put $f(x) = \frac{2x+3}{x+2}$ then we want to find the limit of $f^{(n)}(1)$ the $n$th iterate of $f$.
If we show that $f$ contracts some interval containing $1$ we have proved existence of the limit by Banach Fixed Point Theorem, i.e. we need to show $|f(a)-f(b)|<\eta^{-1}|a-b|$ for $a,b \in [1,2]$ and some $\eta^{-1} < 1$. Since $|f(a)-f(b)| = \frac{|a - b|}{|ab + 2a + 2b + 4|}$ we just need to show the denominator $> 4 = \eta$, say, which is obviously true.
To calculate what the limit is is equivalent to solving $f(x) = x$ or $\frac{x^2 - 3}{x + 2} = 0$ so it's just $\sqrt{3}$.
Alternatively just take the eigenvalues of the matrix [2,3;1,2] to get the fixed points.