Let $a_{n+1} = \frac{4}{7}a_n + \frac{3}{7}a_{n-1}$ and $a_0 = 1$, $a_1 = 2$. Find $\lim_{n \to \infty} a_n$.
1. **Formulate the characteristic equation:**
   The given recurrence relation is:
   \[
   a_{n+1} = \frac{4}{7}a_n + \frac{3}{7}a_{n-1}
   \]
   To solve this, we assume a solution of the form \(a_n = r^n\). Substituting \(a_n = r^n\) into the recurrence relation, we get:
   \[
   r^{n+1} = \frac{4}{7}r^n + \frac{3}{7}r^{n-1}
   \]
   Dividing through by \(r^{n-1}\), we obtain the characteristic equation:
   \[
   r^2 = \frac{4}{7}r + \frac{3}{7}
   \]
   Multiplying through by 7 to clear the fractions, we get:
   \[
   7r^2 = 4r + 3
   \]
   Rearranging terms, we obtain the characteristic polynomial:
   \[
   7r^2 - 4r - 3 = 0
   \]

2. **Solve the characteristic equation:**
   We solve the quadratic equation \(7r^2 - 4r - 3 = 0\) using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 7\), \(b = -4\), and \(c = -3\):
   \[
   r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 7 \cdot (-3)}}{2 \cdot 7}
   \]
   \[
   r = \frac{4 \pm \sqrt{16 + 84}}{14}
   \]
   \[
   r = \frac{4 \pm \sqrt{100}}{14}
   \]
   \[
   r = \frac{4 \pm 10}{14}
   \]
   This gives us two roots:
   \[
   r_1 = \frac{14}{14} = 1 \quad \text{and} \quad r_2 = \frac{-6}{14} = -\frac{3}{7}
   \]

3. **Form the general solution:**
   The general solution to the recurrence relation is:
   \[
   a_n = A(1)^n + B\left(-\frac{3}{7}\right)^n
   \]
   Simplifying, we get:
   \[
   a_n = A + B\left(-\frac{3}{7}\right)^n
   \]

4. **Determine the constants \(A\) and \(B\):**
   Using the initial conditions \(a_0 = 1\) and \(a_1 = 2\):
   - For \(a_0 = 1\):
     \[
     A + B = 1
     \]
   - For \(a_1 = 2\):
     \[
     A + B\left(-\frac{3}{7}\right) = 2
     \]

   We now solve this system of equations:
   \[
   \begin{cases}
   A + B = 1 \\
   A - \frac{3}{7}B = 2
   \end{cases}
   \]

   Subtract the first equation from the second:
   \[
   \left(A - \frac{3}{7}B\right) - (A + B) = 2 - 1
   \]
   \[
   -\frac{10}{7}B = 1
   \]
   \[
   B = -\frac{7}{10}
   \]

   Substitute \(B = -\frac{7}{10}\) back into \(A + B = 1\):
   \[
   A - \frac{7}{10} = 1
   \]
   \[
   A = 1 + \frac{7}{10} = \frac{17}{10}
   \]

5. **Find the limit as \(n \to \infty\):**
   The general solution is:
   \[
   a_n = \frac{17}{10} + \left(-\frac{7}{10}\right)\left(-\frac{3}{7}\right)^n
   \]
   As \(n \to \infty\), \(\left(-\frac{3}{7}\right)^n \to 0\) because \(\left|-\frac{3}{7}\right| < 1\). Therefore:
   \[
   \lim_{n \to \infty} a_n = \frac{17}{10}
   \]

The final answer is \(\boxed{\frac{17}{10}}\).