Sequences $(x_n)_{n\ge1}$, $(y_n)_{n\ge1}$ satisfy the relations $x_n=4x_{n-1}+3y_{n-1}$ and $y_n=2x_{n-1}+3y_{n-1}$ for $n\ge1$. If $x_1=y_1=5$ find $x_n$ and $y_n$. 

Calculate $\lim_{n\rightarrow\infty}\frac{x_n}{y_n}$.
1. **Formulate the system of recurrence relations:**
   Given the sequences \((x_n)_{n \ge 1}\) and \((y_n)_{n \ge 1}\) with the relations:
   \[
   x_n = 4x_{n-1} + 3y_{n-1}
   \]
   \[
   y_n = 2x_{n-1} + 3y_{n-1}
   \]
   and initial conditions \(x_1 = y_1 = 5\).

2. **Set up the characteristic equation:**
   To solve the system, we use the characteristic equation method. We write the system in matrix form:
   \[
   \begin{pmatrix}
   x_n \\
   y_n
   \end{pmatrix}
   =
   \begin{pmatrix}
   4 & 3 \\
   2 & 3
   \end{pmatrix}
   \begin{pmatrix}
   x_{n-1} \\
   y_{n-1}
   \end{pmatrix}
   \]
   Let \(A = \begin{pmatrix} 4 & 3 \\ 2 & 3 \end{pmatrix}\). The characteristic polynomial of \(A\) is given by:
   \[
   \det(A - \lambda I) = \det \begin{pmatrix} 4 - \lambda & 3 \\ 2 & 3 - \lambda \end{pmatrix} = (4 - \lambda)(3 - \lambda) - 6 = \lambda^2 - 7\lambda + 6
   \]
   Solving the characteristic equation \(\lambda^2 - 7\lambda + 6 = 0\), we get:
   \[
   \lambda = \frac{7 \pm \sqrt{49 - 24}}{2} = \frac{7 \pm 5}{2}
   \]
   Thus, the eigenvalues are \(\lambda_1 = 6\) and \(\lambda_2 = 1\).

3. **Find the eigenvectors:**
   For \(\lambda_1 = 6\):
   \[
   (A - 6I) \mathbf{v} = 0 \implies \begin{pmatrix} -2 & 3 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0
   \]
   Solving, we get \(v_1 = 3k\) and \(v_2 = 2k\). Thus, an eigenvector corresponding to \(\lambda_1 = 6\) is \(\mathbf{v}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\).

   For \(\lambda_2 = 1\):
   \[
   (A - I) \mathbf{v} = 0 \implies \begin{pmatrix} 3 & 3 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = 0
   \]
   Solving, we get \(v_1 = -v_2\). Thus, an eigenvector corresponding to \(\lambda_2 = 1\) is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

4. **Form the general solution:**
   The general solution to the system is:
   \[
   \begin{pmatrix}
   x_n \\
   y_n
   \end{pmatrix}
   = c_1 6^n \begin{pmatrix} 3 \\ 2 \end{pmatrix} + c_2 1^n \begin{pmatrix} 1 \\ -1 \end{pmatrix}
   \]
   Simplifying, we get:
   \[
   x_n = 3c_1 6^n + c_2
   \]
   \[
   y_n = 2c_1 6^n - c_2
   \]

5. **Determine the constants \(c_1\) and \(c_2\):**
   Using the initial conditions \(x_1 = y_1 = 5\):
   \[
   x_1 = 3c_1 6 + c_2 = 18c_1 + c_2 = 5
   \]
   \[
   y_1 = 2c_1 6 - c_2 = 12c_1 - c_2 = 5
   \]
   Solving these equations simultaneously:
   \[
   18c_1 + c_2 = 5
   \]
   \[
   12c_1 - c_2 = 5
   \]
   Adding these equations:
   \[
   30c_1 = 10 \implies c_1 = \frac{1}{3}
   \]
   Substituting \(c_1\) back into the first equation:
   \[
   18 \left(\frac{1}{3}\right) + c_2 = 5 \implies 6 + c_2 = 5 \implies c_2 = -1
   \]

6. **Write the explicit formulas for \(x_n\) and \(y_n\):**
   \[
   x_n = 3 \left(\frac{1}{3}\right) 6^n - 1 = 6^n - 1
   \]
   \[
   y_n = 2 \left(\frac{1}{3}\right) 6^n + 1 = 4 \cdot 6^{n-1} + 1
   \]

7. **Calculate the limit:**
   \[
   \lim_{n \to \infty} \frac{x_n}{y_n} = \lim_{n \to \infty} \frac{6^n - 1}{4 \cdot 6^{n-1} + 1} = \lim_{n \to \infty} \frac{6^n}{4 \cdot 6^{n-1}} = \lim_{n \to \infty} \frac{6^n}{4 \cdot 6^{n-1}} = \lim_{n \to \infty} \frac{6 \cdot 6^{n-1}}{4 \cdot 6^{n-1}} = \frac{6}{4} = \frac{3}{2}
   \]

The final answer is \( \boxed{ \frac{3}{2} } \)