Let $f : \mathbb{R}\to\mathbb{R}_+$ be a smooth function. Given some $x_0\in\mathbb{R}$ and a direction $s\in\{-1,1\}$, I'm interested in image of $x_0$ under $T_n$ composed with itself $n$ times, i.e. ${T_n}^n(x_0)$ where
$$T_n:x\mapsto x+f(x)s/n$$
If the limit as $n\to\infty$ exists, how to find it? Is there some integral-equation or a differential equation that represents the limit behavior?
E.g: it seems to converge when moving forward from $x_0=0$ with $f(x) = \sqrt{1+x^2}$:
${T_n}^n(x_0)=1.0886792,\ \text{for }n=3$
${T_n}^n(x_0)=1.1643523,\ \text{for }n=30$
${T_n}^n(x_0)=1.1740887,\ \text{for }n=300$
${T_n}^n(x_0)=1.1750897,\ \text{for }n=3000$
${T_n}^n(x_0)=1.1751900,\ \text{for }n=30000$
${T_n}^n(x_0)=1.1752001,\ \text{for }n=300000$
The limit in your example there is $\sinh(1)=\frac{e-e^{-1}}{2}\approx 1.175201194$.
In more generality, we can solve a problem like this by transforming it into a differential equation. The first-order autonomous differential equation $\frac{dx}{dt}=f(x)$, for $t$ from $0$ to $1$ with $x(0)=x_0$, gives $x(1)\approx T_n^n(x_0)$ by Euler's method with $n$ equally-sized steps. Well, that's $s=1$. The alternative $s=-1$ would be the approximation to $x(-1)$.
If the differential equation has a solution on that interval, then Euler's method converges to the solution - in theory, at least. In practice, convergence is slow enough that roundoff error will become a problem if we try to use enough steps for high accuracy.
Solving such equations is fairly straightforward. I'll demonstrate with your example: $$\frac{dx}{dt}=\sqrt{1+x^2}$$ $$\int\frac1{\sqrt{1+x^2}}\,dx = \int dt$$ $$\ln(x+\sqrt{1+x^2}) = \sinh^{-1}(x) = t+c$$ $$x=\sinh(t+c)$$ Then, from $x(0)=0$, we get $x=\sinh t$, and $x(1)=\sinh 1$. The key step in the solution there, applicable to any first-order autonomous equation, is to treat the $dx$ and $dt$ as if they were variables we could move around, and separate them so we can integrate. Differential equations of this sort are also called separable because of that method.
Because of the form here, the integral on the $t$ side will always just be $t$; everything that changes is on the $x$ side, and integrating gives us an implicit solution with $t$ written in terms of $x$. We then have to invert the function (easier said than done) to write $x$ in terms of $t$.
Now, that bit I said about "if the equation has a solution on that interval"? No guarantees there. If we tried this for $f(x)=1+4x^2$ (and $x_0=0$) without the square root, we would get divergent behavior since the solution $x=\tan 2t$ to the differential equation blows up to $\infty$ at $\frac{\pi}{4}$.