Correct calculation of recursive limit

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 $\lim\limits_{\delta t \rightarrow 0} x_{i+1} = x_0$ ?

Let $n$ be an integer. I believe what you mean to express is the sequence of functions $\{x^{(n)}\}$ with
\begin{gather*} x^{(n)}((k+1)\delta t)=x^{(n)}(k\delta t)\exp(\mu\delta t+Y_{k}^{(n)}\sigma(x^{(n)}(k\delta t))\sqrt{\delta t})\\ \text{where }\delta t=T/n\text{ and }0\leq k<n\text{ is an integer}. \end{gather*} $Y_{k}^{(n)}$ denotes a random variable with support $\{-1,+1\}$. $x^{(n)}(0)$ is a constant independent of $n$.
When $t/\delta t$ is not an integer, $x^{(n)}(t)$ is defined by linear interpolation. The limit you care about is $$\lim_{n\rightarrow\infty}x^{(n)}(t).$$ Note that $n$ depends on $\delta t$. Another subtlety is that this is not a limit in the usual sense, since randomness is involved (look up "convergence in distribution").