1D Ising model thermodynamic limit The partition function for the infinite range Ising model is 
$$Z(\beta ,h)=\int_{-\infty}^{\infty}dm \frac{1}{\sqrt{2\pi/N\beta J}} e^{-Ng(m)}$$
where $g(m) = \frac{\beta J}{2}m^2 - \ln\left[2\cosh(\beta h + \beta J m)\right]$. Here $J$ is the coupling strength and $h$ is the applied field. To get the analytical partition function one then takes the thermodynamic limit $N\rightarrow \infty$. The magnetization is then defined by
$$M = \lim_{h\rightarrow 0} \lim_{N\rightarrow\infty} \frac{1}{\beta N} \frac{\partial \ln Z(\beta,h)}{\partial h}.$$
My question is then following. Why is it important to take the $N\rightarrow\infty$ first and then the $h\rightarrow0$ limit. If we take instead the following quantity
$$M_1 =  \lim_{N\rightarrow\infty} \lim_{h\rightarrow 0} \frac{1}{\beta N} \frac{\partial \ln Z(\beta,h)}{\partial h},$$
why wouldn't $M_1$ be a valid quantity for magnetization? Since we are taking the $N\rightarrow \infty$ still one should be able to perform the integration by steepest descent method. What stops it to be the magnetization of the system? 
 A: This is an issue of limits not commuting which applies to any version of the Ising model with a phase transition, not just the infinite-range case. I will give a rough version of the explanation found in chapter 2 of Nigel Goldenfeld's book.
For any finite $N$, the partition function is a sum of a finite number of analytic functions (exponentials). Thus, for any finite $N$ the magnetization must be a smooth function, and in particular be continuous. At $h = 0$, symmetry requires that the magnetization $M = 0$. Thus, by this continuity, as $h\to0$ from either side, $M\to0$ smoothly.  Therefore, if we take the limit $h \to 0$ at finite $N$ before taking the limit $N \to 0$, we will always 
find that there is no "spontaneous magnetization" at $h \approx 0$.
On the other hand, the limit of an infinite sum of continous functions need not be continous (consider e.g. the Fourier representation of a square wave).  Therefore, taking the limit $N\to \infty$ first allows us to get an $M{\left(h\right)}$ which is not continous at $h = 0$. So although symmetry still requires that $M{\left(0\right)} = 0$, we can have $\lim_{h\to0^+} M{\left(h\right)} > 0$.
This figure from Nigel's book illustrates this discontinuity:

