Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as:
$u_1=6,u_{n+1} = \frac{2n+a}{n} + \sqrt{\frac{n+a}{n}u_n+4}, \forall n \ge 1$
a) With $a=0$, prove that there exist a finite limit of $(u_n)$ and find that limit
b) With $a \ge 0$, prove that there exist a finite limit of $(u_n)$
### Part (a)
Given \( a = 0 \), we need to prove that there exists a finite limit of \( (u_n) \) and find that limit.

1. **Initial Condition and Recurrence Relation:**
   \[
   u_1 = 6, \quad u_{n+1} = \frac{2n}{n} + \sqrt{\frac{n}{n}u_n + 4} = 2 + \sqrt{u_n + 4}
   \]

2. **Establishing a Lower Bound:**
   \[
   u_n > 5 \quad \forall n \ge 1
   \]
   This is because \( u_1 = 6 \) and the recurrence relation ensures that \( u_n \) remains above 5.

3. **Analyzing the Difference:**
   \[
   |u_{n+1} - 5| = \left| 2 + \sqrt{u_n + 4} - 5 \right| = \left| \sqrt{u_n + 4} - 3 \right|
   \]

4. **Bounding the Difference:**
   \[
   \sqrt{u_n + 4} - 3 = \frac{u_n - 5}{\sqrt{u_n + 4} + 3}
   \]
   Since \( u_n > 5 \), we have \( \sqrt{u_n + 4} + 3 > 5 \). Therefore,
   \[
   |u_{n+1} - 5| = \frac{|u_n - 5|}{\sqrt{u_n + 4} + 3} < \frac{1}{5} |u_n - 5|
   \]

5. **Iterative Inequality:**
   \[
   |u_{n+1} - 5| < \frac{1}{5} |u_n - 5| < \left( \frac{1}{5} \right)^n |u_1 - 5|
   \]
   Since \( |u_1 - 5| = 1 \),
   \[
   \lim_{n \to \infty} |u_n - 5| = 0 \implies \lim_{n \to \infty} u_n = 5
   \]

### Part (b)
Given \( a \ge 0 \), we need to prove that there exists a finite limit of \( (u_n) \).

1. **Initial Condition and Recurrence Relation:**
   \[
   u_1 = 6, \quad u_{n+1} = \frac{2n + a}{n} + \sqrt{\frac{n + a}{n}u_n + 4}
   \]

2. **Defining a New Sequence:**
   Let \( v_n = \frac{u_n}{n} \). Then,
   \[
   v_1 = 6, \quad v_{n+1} = \frac{\frac{2n + a}{n} + \sqrt{(n + a)v_n + 4}}{n + 1}
   \]

3. **Simplifying the Recurrence Relation:**
   \[
   v_{n+1} < \frac{2n + a}{n^2 + n} + \frac{\sqrt{(n + a)v_n}}{n + 1} + \frac{2}{n + 1}
   \]
   Using the inequality \( \sqrt{a + b} < \sqrt{a} + \sqrt{b} \),
   \[
   v_{n+1} \le \frac{2n + a}{n^2 + n} + \frac{2}{n + 1} + \frac{1}{2} \left( \frac{1}{n + 1} + \frac{(n + a)v_n}{n + 1} \right)
   \]

4. **Applying the Arithmetic Mean-Geometric Mean Inequality (AM-GM):**
   \[
   v_{n+1} \le \frac{2n + a}{n^2 + n} + \frac{5}{2(n + 1)} + \frac{1}{2} \cdot \frac{n + a}{n + 1} v_n
   \]

5. **Bounding the Terms:**
   Since \( \lim_{n \to \infty} \frac{n + a}{n + 1} = 1 \), there exists \( N \) such that for all \( n \ge N \),
   \[
   \frac{n + a}{n + 1} < \frac{3}{2}
   \]
   Therefore, for \( n \ge N \),
   \[
   v_{n+1} < \left( \frac{2n + a}{n^2 + n} + \frac{5}{2(n + 1)} \right) + \frac{3}{4} v_n
   \]

6. **Applying the Lemma:**
   By the well-known lemma, if \( z_n \) is a positive sequence and \( z_{n+1} \le cz_n + q_n \) with \( \lim q_n = 0 \) and \( c \in (0, 1) \), then \( \lim z_n = 0 \). Here, \( v_n \) satisfies this condition, so
   \[
   \lim_{n \to \infty} \frac{u_n}{n} = 0
   \]

7. **Analyzing the Difference Again:**
   \[
   |u_{n+1} - 5| = \left| \frac{u_n - 5 + \frac{au_n}{n}}{\sqrt{\frac{n + a}{n} u_n + 4} + 3} + \frac{a}{n} \right| < \frac{1}{5} |u_n - 5| + \left( \frac{a}{5} \cdot \frac{u_n}{n} + \frac{a}{n} \right)
   \]

8. **Applying the Lemma Again:**
   Since \( \lim_{n \to \infty} \frac{u_n}{n} = 0 \) and \( \frac{a}{n} \to 0 \), applying the lemma, we have
   \[
   \lim_{n \to \infty} |u_n - 5| = 0 \implies \lim_{n \to \infty} u_n = 5
   \]

The final answer is \( \boxed{ 5 } \)