$(\frac{x}{a})^n + (\frac{y}{b})^n = 1$ as $n \to \infty$ To analyse what happens to the function $$\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n = 1$$ when as $n \to \infty$.
Here we are given, $a=4,b=10$. When $n=1,2$ I know it is a straight line and ellispe. What happens when $n \to \infty$?
I have plotted it and saw that it converges to rectangle, but can't explain it mathematically!
 A: I think intuitively you could think like this.
There is an assumption that $-a\le x\le a$ and $-b \le y \le b$.
Transform to new coordinates $x'=x/a$ and $y'=y/b$ leading to
$$(x')^n + (y')^n = 1$$
$$-1 \le x' \le 1, -1 \le y' \le 1$$
For all $|x'| \lt1$
$$
\lim x'\to 0, n \to\infty
$$
Similarly $y'$.
The only non-zero limits are for $x'=\pm1$ and $y'=\pm1$ i.e. when $x=\pm a$ and $y=\pm b$
A: Yes, it converges to the rectangle in the following sense: 
Let $p = (x, y)$ be a point in the curve $\left( \frac xa \right)^n + \left(\frac yb\right)^n = 1$. 
Then either $\left( \frac xa\right)^n\ge \frac 12$ or $\left( \frac yb\right)^n \ge \frac 12$, which is 
$$ \left|x\right| \ge \sqrt[n]{\frac 12}|a| \ \ \text{or }  \left|y\right| \ge \sqrt[n]{\frac 12}|b|. $$
Thus the distance between $p$ and the rectangle $\{ |x| = |a|, |y|= |b|\}$ is less then 
$$ \left( 1- \sqrt[n]{\frac 12}\right) \min\{|a|, |b|\}\to 0$$
as $n\to \infty$. In the above sense the curve converges to the rectangle as $n\to \infty$. 
One rigorous way is to use the Hausdorff distance between the curve and rectangle (as subsets in $\mathbb R^2$) and show that the Hausdorff disatnce tends to zero.
