Help evaluating $\lim_{n \to \infty}(x^n + y^n)^{1/(1-n)}$ I get $0$ when I do it by hand, but when i check mathematica I get $\frac{1}{\max\{x,y\}}$ (at least for the 4 or 5 cases I have checked so far).
Here is my work: Take natural log to get $\frac{1}{1-n} \ln(x^n + y^n)$, then L'hopital on that (I suspect this may be where I am wrong, as perhaps L'hopital doesn't apply?), giving $-\frac{n(x^{n-1}+y^{n-1})}{x^n + y^n}$
That then simplifies to $-n$, and $\lim_{n\to\infty} -n = -\infty$, and $e^{-\infty} =0$
So how can I go about evaluating this limit? Specifically, since I do think that the answer should be $1/\max\{x,y\}$, how does $\max$ get involved?
Thanks.

Consider $$A_n=(x^n + y^n)^{\frac 1 {1-n}}$$  suppose $x>y$ and set $y=\lambda x$ with $\lambda <1$.
So $$A_n=x^{\frac{n}{1-n}}(1+\lambda^n )^{\frac{1}{1-n}}$$ So, when $n$ becomes large, the first exponent tends to $-1$. Concerning the second term, since $n$ is large $\lambda^n\to 0$ and raised to a power going to $0$, it tends to $1$. All of this makes the limit to be $\frac 1x$.
Repeat the same process assuming now $y>x$ and get a limit equal to $\frac 1y$.
$$\lim_{n \to \infty}(x^n + y^n)^{\frac 1 {1-n}}=\frac{1}{\max\{x,y\}}$$
The fat that $x+y=1$ is not relevant.
