Simple Calculus Proof I'm having some difficulty with this proof. Could you help me?

Let $a,b>0$ and $c_n = (a^n + b^n)^{1/n}$. Show that $$\lim_{n\to \infty}c_n = \max \{a,b\}.$$

Thank you.
 A: Suppose that $b \ge a$.  Then $$c_n = b ((a/b)^n + 1)^{1/n}.$$ Since $$1 \le ((a/b)^n + 1)^{1/n} \le 2^{1/n}$$ and $2^{1/n} \to 1$ as $n \to \infty$ you can use the squeeze theorem to obtain the limit.
A: Hint: factor out the biggest of the two; you will get $\max(a,b)\cdot x_n$, where $x_n\to 1$.



*

*If $a=b$, then
$$
(a^n+b^n)^{1/n} = (2a^n)^{1/n} = 2^{1/n} a \xrightarrow[n\to\infty]{} 1\cdot a = a.
$$

*If $a\neq b$, suppose (without loss of generality) that $a > b$.
$$
(a^n+b^n)^{1/n} = \left(a^n\left(1+\frac{b^n}{a^n}\right)\right)^{1/n} = a\left(1+\frac{b^n}{a^n}\right)^{1/n} \xrightarrow[n\to\infty]{} a\cdot 1 = a.
$$
(using the fact that $0<\frac{b}{a}< 1$)
A: There is a much shorter method than the one below. Let me present the idea.
If $b>a$:
$$\sqrt[n]{a^n+b^n}=b\sqrt[n]{\left(\frac ab\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=b \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ab\right)^n+1}$$
$$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= b\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {a^n}{b^n}\right)+1}=b$$
If $a>b$:
$$\sqrt[n]{a^n+b^n}=a\sqrt[n]{\left(\frac ba\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ba\right)^n+1}$$
$$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= a\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {b^n}{a^n}\right)+1}=a$$
If $a=b$:
$$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a\lim_{n \rightarrow \infty}2^{1/n}=a2^0=a=b$$
In either case we can see that the maximum of $a,b$ happens to be the limit. Thus:
$$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= \max (a,b)\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {(\min(a,b))^n}{(\max(a,b))^n}\right)+1}=\max (a,b)$$
