Preference relations defined by $x_1^n + x_2^n$ converge to $\max\{x_1, x_2\}$ In the problem set 2 of Rubinsteins Microeconomics (btw is there a comparably nice written book on macroeconomics?) there is the following question:
Let $\succ_n$ be the preference relations defined on $\mathbb{R}^2_+$ by the utility $x_1^n + x_2^n$. Let the preference relations $\succ$ be defined by the utility $\max\{x_1, x_2\}$. Show that $\succ_n$ converge to $\succ$.

Preference relations are said to converge if for $a \succ b$ we have that $a \succ_m b$ for sufficiently large $m$. I am indeed able to prove that.

Now w.l.o.g. $x_1 = \max\{x_1, x_2\}$ and $y_1 = \max\{y_1, y_2\}$. Assuming that $x_1 = y_1$ we have $x \sim y$. But the only case when $x \sim_m y$ is when also $x_2 = y_2$. In other cases we would have either $x \succ_m y$ or $y \succ_m x$ for all $m$ (depending on $x_2$ and $y_2$, the "smaller component". These are actually lexicographic preferences!)

Is there a reason that the convergence of preference relations is done in this way to ignore such subtilities? In my opinion we should have that $\succeq = \lim_{n \to \infty} \succeq_n$ iff $a \succeq b = \lim_{n \to \infty} a \succeq_n b$.
I am assuming you're familiar with the concept of monotonic transformations.

Let $f(z)=z^n$ be a monotonic transformation. $f(z)$ is an order preserving transformation for all positive $n$. Let $v:\mathbb{R_+^2} \to\mathbb{R}$ be an utility function, where $v(x_1,x_2)=(x_1^n + x_2^n)^{1/n}$. The function $f(v(x_1,x_2))=((x_1^n + x_2^n)^{1/n})^n=x_1^n + x_2^n$ will represent the same preferences as $v(x_1,x_2)$. Therefore, $(x_1^n + x_2^n)^{1/n}$ should also converge to $\max\{x_1,x_2\}$. 

To prove the same, consider $v(x_1,x_2)=\Big(1+\cfrac{x_1^n }{x_2^n}\Big)^{1/n}x_2$. Assume $x_2 = \max\{x_1,x_2\}. $Now, as $n \to \infty$, two cases arise. 

1) If $x_2&gt;x_1$, then $\cfrac{x_1^n }{x_2^n} \to 0$ as $n \to \infty$. In that case, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.

2) If $x_2=x_1$, the limit can be calculated by taking log on both sides as follows-
\begin{align*}
\ln(v(x_1,x_2)) = \cfrac{\ln(1+\cfrac{x_1^n }{x_2^n})}{n} +\ln(x_2) \,\,\,\,-(1)
\end{align*}
Since $x_1/x_2=1$,we have $(x_1/x_2)^n=1^n$. Using this in equation $(1)$, we get
\begin{align*}
\ln(v(x_1,x_2)) = \cfrac{\ln(1+1^n)}{n} +\ln(x_2) 
\end{align*}
As $n\to \infty$, the term $\cfrac{\ln(1+1^n)}{n}$ becomes $\infty/\infty$. This limit can be solved using L'Hospital's rule. The value of the limit(you may check it for yourself) comes out to be $0$.
Now we have $\ln(v(x_1,x_2)) = 0 +\ln(x_2)$, or, $v(x_1,x_2)=x_2=\max\{x_1,x_2\}$.