Solution: We will first claim that $(x_n)$ is an decreasing sequence. Note that \begin{align*} x_n - x_{n + 1} & = \frac{1}{3} \left( x_{n-1} + 1 \right) - \frac{1}{3} \left( x_{n} + 1 \right) \\ & = \frac{1}{3}\left( x_{n-1} - x_n \right) \\ & = \frac{1}{3} \left( \frac{1}{3} \left( x_{n-2} - x_{n-1} \right) \right) = \left( \frac{1}{3} \right) ^2 \left( x_{n-2} - x_{n-1} \right) \\ & = \cdots \cdots \\ & = \left( \frac{1}{3} \right) ^{n-1} \left( x_1 - x_2 \right) = \left(\frac{1}{3}\right)^{n-1} \left( 1 - \frac{2}{3} \right) \geq 0. \end{align*} Thus, $(x_n)$ is a decreasing sequence.

It is clear that the sequence $(x_n)$ is bounded below by $0$. Therefore, the sequence will converge (monotone convergence theorem). Let the limit of the sequence be $l$. Then $x_n \to l$ which implies $x_{n-1} \to l $. Thus, \begin{align*} & \lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{1}{3}\left( x_{n-1} + 1 \right) \\ \implies & l = \frac{1}{3} (l + 1) \\ \implies & l = \frac{1}{2}. \end{align*} Thus, the sequence is convergent and the lim is $\frac{1}{2}$.