Prove it. How do I start?

Question

Given $a,b >0$

$$ \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} = max\{a,b\}$$

I tried to start with taking a common but could reach the conclusion , please provide a solution.

Answer 1

If $a\leq b$, then $b=\sqrt[n]{b^n}\leq\sqrt[n]{a^n+b^n}\leq\sqrt[n]{b^n+b^n}=\sqrt[n]{2b^n}=\sqrt[n]{2}b$. Since the right hand side goes to $b$, the result follows from the sandwich principle (squeezing theorem). For $a>b$ the proof is similar.

Answer 2

Intuitively, as $n \to \infty$, if $a>b$ then $a^n$ dominates the $b^n$ term as $n \to \infty$. So $\lim \limits_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} \sim \lim \limits_{n \to \infty} (a^n)^{\frac{1}{n}} = a.$

If $a=b$, then we have $\lim \limits_{n \to \infty} (2a^n)^{\frac{1}{n}} = \lim \limits_{n \to \infty} \sqrt[n]{2} a = a.$

Answer 3

Let $a > b$. You have $$\lim_{n\rightarrow\infty}\sqrt[n]{a^n+b^n} = \lim_{n\rightarrow\infty} a \underbrace{\sqrt[n]{1+\frac{b}{a}^n}}_{\rightarrow\ 1} = a$$

Answer 4

Hint:$$a^n+b^n=a^n\left(1+\frac{b^n}{a^n}\right)=b^n\left(1+\frac{a^n}{b^n}\right). $$

Answer 5

If $a \leq b$, then $$ 1^{\frac{1}{n}} \leq \left( \left( \frac{a}{b}\right)^n +1 \right)^{\frac{1}{n}} \leq (1^n +1)^{\frac{1}{n}} = 2^{\frac{1}{n}}$$ So, by the squeeze theorem,

$$\lim_n \left( \left( \frac{a}{b}\right)^n +1 \right)^{\frac{1}{n}} = 1 $$

multiplying by $b$ both sides, you get

$$\lim_n (a^n + b^n)^{\frac{1}{n}} = b$$

Answer 6

Let us observe that: $$\max\{a,b\}=(0+\max\{a^n,b^n\})^{\frac{1}{n}}\le (a^n+b^n)^{\frac{1}{n}}\le (\max\{a^n,b^n\}+\max\{a^n,b^n\})^{\frac{1}{n}}\le (2\max\{a^n,b^n\})^{\frac{1}{n}}=\sqrt[n]{2}\max\{a,b\}$$ Now, let us calculate limits (as $n\to \infty$) of both sides: $$\lim\limits_{n\to \infty} \max\{a,b\}=\max\{a,b\}$$ $$\lim\limits_{n\to \infty} \big(\sqrt[n]{2}\max\{a,b\}\big)=\max\{a,b\}\cdot \lim\limits_{n\to \infty} \sqrt[n]{2}=\max\{a,b\}$$ So by the squeeze theorem we have: $$\lim\limits_{n\to \infty} \left(a^n+b^n\right)^{\frac{1}{n}}=\max\{a,b\}$$