Solution
= Solution
Since $0<a\leq b$, we have $b^n\leq a^n+b^n\leq2b^n$. Taking positive $n$th roots yields
$$
b\leq(a^n+b^n)^{1/n}\leq b\,2^{1/n}.
$$
For every $\varepsilon>0$, the <binomial theorem> gives $(1+\varepsilon)^n\geq1+n\varepsilon>2$ eventually, so $2^{1/n}\to1$. The <squeeze theorem> therefore gives
$$
\boxed{(a^n+b^n)^{1/n}\longrightarrow b.}
$$