How would one prove, or disprove, that as $n$ goes to $0$, the curve described by $x^n + y^n = 2$ approaches that of $xy = 1$? I was playing around with $n$-norms a few years ago, and wondered, if the limit as $n\rightarrow\infty$ of the curve given by $x^n + y^n = 2$ was equivalent to $max(x,y) = 1$, how the curve would behave as we approached different values of $n$, such as zero or $-\infty$. Interestingly enough, the limit as it approaches $-\infty$ seems to be $min(x,y) = 1$, which was easy enough to guess based on observation, and even more interestingly, the limit as it approaches $0$ appears to be $xy=1$, which I also guessed by simple observation.
How would one go about proving or disproving these conjectures? (For those who are interested, I made an animation of the curve as $n$ varies from $-\infty$ to $\infty$, which is available here: https://youtu.be/WwrsmpjU1Y0)
Edit: I should have mentioned, I'm only considering the curve for $x,y\geq 0$.
 A: Yes, your conjecture that the curve $x^n+y^n=2$ tends to $xy=1$ as $n\to 0$ is correct. Note that $x^n+y^n=2$ if and only if $y=(2-x^n)^{1/n}$. The question at hand here is whether or not this is the same as $1/x$ as $n\to 0$. This is equivalent to the question of whether or not the limit of their ratios is $1$. Well,
$$\begin{split}
\lim_{n\to0}\frac{(2-x^n)^{1/n}}{1/x}&=\lim_{n\to\infty}x\left(2-x^{1/n}\right)^n\\&=x\underbrace{\lim_{n\to\infty}\left(2-x^{1/n}\right)^n}_{L(x)}.
\end{split}$$
It turns out that this limit $L(x)$ is actually $1/x$, so the result is indeed true. To show this, take the logarithm and write the limit as
$$\log L(x)=\lim_{n\to\infty}\frac{\log(2-x^{1/n})}{1/n}, $$
then employ L'Hospital's rule to show that $\log L(x)=-\log x$. I'll leave this to you.
A: The first conjecture is false. For even integers $n>10$ and $x > 1.5$ there is no $y$ satisfying $x^n+y^n < 2$.
So is the 2nd conjecture. For even integers $n > 10$ and any $x>2$, the value of $y$ cannot be any larger than 1. 
Your conjectures would be true for curve
$\left(x^n+y^n\right)^{\frac{1}{n}} = 2$ though [restricted to $x,y >0$]. We show the first modified conjecture. Assuming WLOG that $x \ge y$, we show that $x$ must approach 2 from below. Indeed, for positive $n$ the string of inequalities hold:
$$x = (x^n)^{\frac{1}{n}} \le  2 = \left(x^n+y^n\right)^{1/n} \le (2x^{n})^{1/n} \le x(1+2/n)$$
How would you show the modified second conjecture.
