TITLE: How to prove that $\lim\limits_{n \to \infty} s_n = s$ if $a_n^{s_n} + b_n^{s_n} =1$?
QUESTION [6 upvotes]: Suppose $${a_n^{s_n} + b_n^{s_n} =1}$$  $$\lim_{n \to \infty} a_n = a >0$$  $$\lim_{n \to \infty} b_n = b >0$$ and $${a^s + b^s =1}$$ where $a+b<1$. Then how can we prove that $\displaystyle\lim_{n \to \infty} s_n = s$?

REPLY [3 votes]: Consider the function
$$f(x,y,t):=x^t+y^t-1\quad(0<x<1,\ 0<y<1,\ t\in{\mathbb R})\ .$$
By assumption $(a,b,s)\in{\rm dom}(f)$ and $f(a,b,s)=0$. Furthermore
${\partial f\over\partial t}=\log x\ x^t+\log y\ y^t$, whence
$$\left.{\partial f\over\partial t}\right|_{(a,b,s)}=\log a\ a^s+\log b\ b^s<0\ .$$
It follows by the implicit function theorem that there is an open box $W=U\times V\subset{\mathbb R}^3$ with center $(a,b,s)$ and a $C^1$-function $$\psi:\quad U\to V,\quad (x,y)\to t=\psi(x,y)\ ,$$  such that within this box the equation $f(x,y,t)=0$ is equivalent to $t=\psi(x,y)$. 
Coming back to  the original question we now have
$$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \psi(a_n,b_n)=\psi(a,b)=s\ .$$