When can a function $f(x_0 - x)$ be approximated as $f(x_0 - x) = f(x_0) - f'(x_0) x$? When can a function $f(x_0 - x)$ be approximated as $f(x_0 - x) = f(x_0) - f'(x_0) x$? In Reif's statistical mechanics it is said that when $x$ is much smaller than $x_0$ then the approximation can be made. But if $x$ itself is a large number, then the $x^2, x^3, \cdots$ terms will also be important. Why then can we use this linear approximation?
For example, in deriving canonical distribution, $x_0$ is the total energy of the system A and the heat reservoir A'. $x$ is the energy of the system A which is much smaller than $x_0$. Then it is said that $\ln\Omega'(x_0-x)=\ln\Omega'(x_0)-(\partial\ln\Omega'/\partial x')_0 x$. Why?
 A: Typically when we approximate functions via Taylor series in physics, it's always best practice to "non-dimensionalize" your functions first, and expand in a unitless variable. In this case, this would mean converting $f(x-x_0)$ to $f(1-x/x_0)$. That process may itself be nontrivial, but once it's done, the accuracy of the expansion becomes trivial to see.
If for some reason you can't or won't do that, then:
We have to be very careful what we mean when we say that $x$ is much smaller than $x_0$.
This is essentially just the first two terms of a Taylor series centered at $x_0$. You can do this whenever $f$ is locally analytic at $x_0$, to any number of terms that you wish (you can also technically do this for non-smooth functions, but with no guarantee that the series actually converges, so do so at your own risk).
As for why it works when $x$ is much smaller than $x_0$, even when $x$ is itself large: the first-term truncation of the Taylor series is considered to be accurate when $x_0-x\approx x_0$ (for functions that are locally analytic at $x_0$, the error on this approximation is given by Taylor's theorem). This is equivalent to saying that $x$ is much smaller than $x_0$. If it's not obvious to you why this is, consider rewriting $x_0-x$ as $x_0\left(1-\frac{x}{x_0}\right)$. If $x_0\left(1-\frac{x}{x_0}\right)\approx x_0$, then it follows that $1-\frac{x}{x_0}\approx 1$, meaning that $\frac{x}{x_0}\ll 1$.
As long as the radius of convergence of the Taylor series about $x_0$ is nonzero, there will always be some finite region around $x_0$ where the quadratic term is unimportant. This follows directly from the definition of differentiability.
The definition of the derivative is as follows:
$$f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$$
In order for this limit to exist, there must be some value that the ratio on the right-hand side converges to. In other words, there must be some threshold $h_{max}$ such that, for all $h<h_{max}$, changing $h$ only varies the ratio within some tolerance $\epsilon$ (which you can make as small as you like; making $\epsilon$ smaller generally means $h_{max}$ will also get smaller). In the region where $h<h_{max}$, we have that $f(x_0+h)-f(x_0)\approx hf'(x_0)$, or in other words, that $f(x_0+h)\approx f(x_0)+f'(x_0)h$.
So we've shown that such a region must always exist, as long as the function is differentiable in the first place. But there are no guarantees made on the specific width of that region. Indeed, if the higher derivatives of $f$ are much larger than the first derivative, then the region in which this happens is very small. Essentially, the higher derivatives of the function determine what the $\approx$ sign specifically means when we say that $x_0-x\approx x_0$. Fortunately, you don't have to calculate the higher derivatives directly to obtain the width of this region; Taylor's theorem gives you an upper bound on the inaccuracy of your linear approximation, and the width of the region of validity is directly dependent on how much error you can accept.
A: The main difficulty you're running into is that there is no absolute sense in which a quantity with units can be considered "large," especially for the purposes of doing calculus. From the context, I would guess that you have three quantities with units that fit into this hierarchy: $E_0 \gg E \gg kT$, and $x_0 = E_0 / kT$ with $x = E/kT$.
For the purposes of approximating $f(x_0 - x) \approx f(x_0) - f'(x_0)x + \ldots$ the comparison that matters not $x$ to $x_0$. The important part comes from the remainder theorem that says that the following relationship is exact:
$$f(x_0-x) = f(x_0) - f'(x_0)x + \frac{f''(\xi)}{2}x^2,$$
for some $0 < \xi < x$. Therefore, the thing you care about is whether you can show that quadratic term is smaller than the desired error (i.e. $x^2 f''(\xi)/2$ is small no matter where between 0 and $x$ $\xi$ is).
The fact that you're approximating $\ln\Omega$, though, means there's a good chance you're not actually using a Taylor series for $\Omega$, but an asymptotic series (specifically Stirling's formula). That adds additional wrinkles. That particular approximation doesn't want small arguments, it works better the larger its argument is, so $x_0 - x$ would also need to be large, requiring $x_0-x \gg 1$ and, with $x\gg 1$ that implies $x_0 \gg x$.
