Define a sequence $\{x_n\}$ as: $\left\{\begin{aligned}& x_1=3 \\ & x_n=\frac{n+2}{3n}(x_{n-1}+2)\ \ \text{for} \ n\geq 2.\end{aligned}\right.$ 
Prove that this sequence has a finite limit as $n\to+\infty.$ Also determine the limit.
1. **Lemma 1**: \( x_n > 1 + \frac{3}{n} \) for \( n \geq 2 \).

   **Proof**: 
   - For \( n = 2 \), we have:
     \[
     x_2 = \frac{2+2}{3 \cdot 2}(x_1 + 2) = \frac{4}{6}(3 + 2) = \frac{4}{6} \cdot 5 = \frac{20}{6} = \frac{10}{3} > \frac{5}{2} = 1 + \frac{3}{2}
     \]
     Thus, the base case holds.
   - Assume \( x_k > 1 + \frac{3}{k} \) for some \( k \geq 2 \). We need to show \( x_{k+1} > 1 + \frac{3}{k+1} \).
     \[
     x_{k+1} = \frac{k+2+1}{3(k+1)}(x_k + 2) = \frac{k+3}{3(k+1)}(x_k + 2)
     \]
     By the induction hypothesis:
     \[
     x_k > 1 + \frac{3}{k}
     \]
     Therefore:
     \[
     x_k + 2 > 3 + \frac{3}{k}
     \]
     Substituting this into the expression for \( x_{k+1} \):
     \[
     x_{k+1} > \frac{k+3}{3(k+1)} \left( 3 + \frac{3}{k} \right) = \frac{k+3}{k+1} \left( 1 + \frac{1}{k} \right)
     \]
     Simplifying:
     \[
     x_{k+1} > \frac{k+3}{k+1} \cdot \frac{k+1+1}{k} = \frac{k+3}{k} = 1 + \frac{3}{k+1}
     \]
     Hence, by induction, \( x_n > 1 + \frac{3}{n} \) for all \( n \geq 2 \).

2. **Lemma 2**: \( x_n < x_{n-1} \) for \( n \geq 3 \).

   **Proof**:
   - By Lemma 1, we have \( x_n > 1 + \frac{3}{n} \).
   - Consider the recurrence relation:
     \[
     x_n = \frac{n+2}{3n}(x_{n-1} + 2)
     \]
     We need to show \( x_n < x_{n-1} \):
     \[
     x_n < x_{n-1} \Longleftrightarrow \frac{n+2}{3n}(x_{n-1} + 2) < x_{n-1}
     \]
     Simplifying:
     \[
     \frac{n+2}{3n}(x_{n-1} + 2) < x_{n-1} \Longleftrightarrow (n+2)(x_{n-1} + 2) < 3nx_{n-1}
     \]
     \[
     n x_{n-1} + 2x_{n-1} + 2n + 4 < 3nx_{n-1}
     \]
     \[
     2x_{n-1} + 2n + 4 < 2nx_{n-1}
     \]
     \[
     2n + 4 < 2nx_{n-1} - 2x_{n-1}
     \]
     \[
     2n + 4 < 2x_{n-1}(n - 1)
     \]
     \[
     n + 2 < x_{n-1}(n - 1)
     \]
     By Lemma 1, \( x_{n-1} > 1 + \frac{3}{n-1} \):
     \[
     x_{n-1}(n - 1) > (1 + \frac{3}{n-1})(n - 1) = n + 2
     \]
     Hence, \( x_n < x_{n-1} \) for \( n \geq 3 \).

3. Since \( x_n \) is decreasing for \( n \geq 2 \) and \( x_n > 1 \) for all \( n \) by Lemma 1, the sequence \( \{x_n\} \) is bounded below and monotone decreasing. By the Monotone Convergence Theorem, \( \{x_n\} \) has a finite limit.

4. Let \( L = \lim_{n \to \infty} x_n \). Taking the limit of both sides in the recurrence relation:
   \[
   L = \lim_{n \to \infty} \frac{n+2}{3n}(x_{n-1} + 2)
   \]
   Since \( x_{n-1} \to L \):
   \[
   L = \lim_{n \to \infty} \frac{n+2}{3n}(L + 2) = \frac{1}{3}(L + 2)
   \]
   Solving for \( L \):
   \[
   3L = L + 2
   \]
   \[
   2L = 2
   \]
   \[
   L = 1
   \]

The final answer is \( \boxed{1} \)