I'm working on this problem but can't find an insight about how to proceed.
The statement: Let $0 \lt a \le b$ and $(x_n) = (a^n + b^n)^\frac1n$ then $ \lim_{n \rightarrow \infty} (x_n) = b$
In an example in the textbook prove that $ (c^\frac1n) \rightarrow 1$ using the expression $c^\frac1n = 1 + d_n$ where $d_n \gt 0$ and then the Bernoulli inequality but does not explain well why there there is $d_n$.
Any hint will be appreciated.
UPDATE
I come up with this regarding $2^\frac1nb$:
Fixing $\epsilon \gt 0$ that $2^\frac1nb \lt b + \epsilon$ then $2 \lt (1 + \frac{\epsilon}b)^n$. By Bernoulli inequality we have $(1 + \frac{\epsilon}b)^n \ge 1 + n\frac{\epsilon}b$. We have that $1 < n\frac{\epsilon}b$ when $n \gt \frac{b}{\epsilon}$. Finally for $n \gt N = [\frac{b}{\epsilon}]$ it follows $2 < 1 + n\frac{\epsilon}{b} \le (1 + \frac{\epsilon}{b})^n$ thus $2^\frac1nb \lt b + \epsilon$. Thank you all for your hints.