I'm constructing a binomial (decision) tree with nodes $ x_i $ according to the following recursion:
$ x_{i+1} = \begin{cases} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} & \text{for next up node} \\ x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} & \text{for next down node} \end{cases} $
with
$ \sigma (x_i) = \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right) $
Taking the limit of $x_{i+1}$ for $\delta t \rightarrow 0$, I get:
$ \lim\limits_{\delta t \rightarrow 0} x_{i+1} = \begin{cases} \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t + \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next up node} \\ \lim\limits_{\delta t \rightarrow 0} x_i e^{\mu \delta t - \sigma(x_i)\sqrt{\delta t}} = x_i & \text{for next down node} \end{cases} $
Therefore, $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_i $. Using this limit recursively, I yield: $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0 $. Therefore:
$ \lim\limits_{\delta t \rightarrow 0} \sigma (x_i) = \lim\limits_{\delta t \rightarrow 0} \sigma_0\left( 1 - \alpha \frac{x_i - x_0}{x_0} \right) = \sigma_0\left( 1 - \alpha \frac{x_0 - x_0}{x_0} \right) = \sigma_0 $
Is my notation correct? Is this last expression correct?
Am I allowed to take the limit recursively to yield $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0 $ ?