What is the purpose of the linear approximation $\Delta L = \alpha L_0 \Delta T$? What is the purpose of the linear approximation $\Delta L = \alpha L_0 \Delta T$?
When using this, we run in all kinds of problems. For example, when a material heates up twice by 1K we get $L=L_1+\alpha L_1=L_0+\alpha L_0+\alpha (L_0+\alpha L_0) = L_0+2\alpha L_0+\alpha^2L_0$. 
However when it heats up by 2K we get $L=L_0+2\alpha L_0$. A similiar thing has been discussed in Bug in linear thermal expansion, $L_0$ must be $0$. This is not a duplicate. 
So is the correct formula  $L = e^{\alpha\Delta T} L_0 $? This looks the most logical when looking to a material heating up increasingly many times, since it then resembles the Taylor series of $e^x$ in $\alpha\Delta T$. If so, what is the purpose of the linear approximation?
 A: The formula will be a good approximation for a "reasonable" temperature range.  However, note in your original problem description, that $$L_0$$ is the length of the object at a standard temperature.  If necessary, you will have to calculate this length, and base all of your temperature differences on the standard temperature that corresponds to this length.  Once that is done, the linear relationship that you noted in your question should work well.
A: Let me expand the argument in the comment. The point is that if we have really a coefficient $\alpha(t)$ we should to consider how the error of its approximation compares with the error of the the second term in the exponential expansion. 
It is a mechanism so standard that we never stop to consider it. You could say that the expansion is a function of the starting temperature and length and the final temperature, imagine $L_f=  L(L_i,T_i,T_f)$. We first go to a differences formulation  $L_f = L_i + \Delta L (L_i,\alpha(T_i), T_f-T_i)$, then $L_f - L_i = L_i K(\alpha(T_i),\Delta T$), then $\Delta L = L_i \alpha(T_i) \Delta T$ then finally the totally linear (in $L$ and $\Delta T$) approximation  $\Delta L = \alpha L_i \Delta T$. 
You can go to the tables and consider each step if your practical need call for it. But you should be integrating in temperature and dilatation, and considering all the sources of inaccuracy. Using an exponential, that anyway is only valid if alpha is constant over all the range of temperature, does not simplify calculations and gives a false feeling of precision.  
A: Reading aloud the formula $\ \Delta L = \alpha L_0 \Delta T\ $ that expresses the elongation of a material in function of the variation of the temperature:
$\frac{\Delta L}{\Delta T}$ is the rate of change of the length in function of the temperature change - this is the tangent to the curve L(T) at any specific point $L_0$ . It is also the coefficient of the first term of the series expansion of the correct function.
 The variation has to be proportional to the initial length $L_0$ (in the RHS). This means that a bar with the double of the length will give the double of the expansion, and so on.
The $ \alpha $ is the coefficient of expansion and depends on the material, is there to say: the materials have different responses to heat. For instance: segments of  of clay and of iron of equal length will elongate differently inside a oven.
Thus the  $\ \alpha L_0  $ is the slope of the tangent. The formula is valid for small variations of the temperature.
$\ \frac{\Delta L}{\Delta T}  = \alpha L $, units of $\alpha=\frac{1}{Temp}$
 and
the solution of $\alpha L(T)=L^{'}(T)$
 is  $L(T) = L_0\cdot e^{\alpha\ T}$, see here  
It appears that what we call first approximation is indeed the complete solution.  
add:
clarification of the self-similarity property of this system;
choose the temperature units (ut) such that $alpha=1$, for a unit length bar ($L_0=1$), we have a simplified relation $l(x) = e^{x}$ where x is dimensionless, depending on T  
the elongation due to two x units variation can be calculated in several ways:
one step of 2x       :
     $l=e^2=7.3890560989$
two steps of 1x each :
     $l=l1+l2; l1=e^1; l2=l1*(e^1)=(e^1)*(e^1)=(e^2)=7.3890560989$
etc ,...
The answer to the OP was complete.
After the original question and my answer, I will put that object in context,  loosing heat and spreading it to the ambient as time goes by.    
Imagine that the object was removed at time=0 (length 1, mass 1, etc) from an oven (at high temperature T=1), scale units at will.
Q1 : Can heat reach an 'infinite' distance away of the object ?
A1 : No, because an infinite time is not available.  
Q2 : Can the object reach any arbitrary small size in the future?
A2a : No (naive and obvious).  When the temp reaches 0ºK the object keeps its minimal size.
A2b : YES (elaborated) . Because particles have associated electromagnetic and gravitational fields, a more a more complete answer should be detailed:
 - When the temp approaches 0ºK , the fields associated with the
   particles of the atoms of the object spread to far away of the
   original configuration. Unstoppable. 
 - Because the fields have energy the particles should exhaust its mass in favor of the growing fields and the minimal size is an evolving shrinking quantity.
That is why I'm considering that a clear cause for the observed space expansion is the matter shrinking, instead of an unexplained and acausal space expansion. 
In fact our units of measure are not based on space but matter properties and by consequence in the lab we can not measure a synchronous slow variation of all objects and rulers.
The evolution of the universe, at large, can be explained in this way (in my profile).   
As we saw the above first equation is able to represent the evolution of the universe (substituting the dependence on temperature decrease with time increase).  
