 Solution

Since $0<a\le b$, we have $b^n\le a^n+b^n\le 2b^n$. Taking positive $n$th roots yieldsFor every $\epsilon >0$, the binomial theorem gives $(1+\epsilon {)}^n\ge 1+n\epsilon >2$ eventually, so $2^{1/n}\to 1$. The squeeze theorem therefore gives