limit

AndrewTom

Find the value of $\lim_{n \to \infty} (a^{n} + b^{n})^{\frac{1}{n}}$ where $a \ge b > 0$ .

**Posted:** Sun Nov 01, 2009 5:17 am

CommandoGuard · P versus NP Offline Joined: 28 Feb 2006 Posts: 33

$a\le\left(a^{n}+b^{n}\right)^{1/n}\le\left(2a^{n}\right)^{1/n}=2^{1/n}a$ . Now apply the squeeze theorem.

**Posted:** Sun Nov 01, 2009 5:28 am

J.Y.Choi

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

Because $1<\left(1+\left(\frac{b}{a}\right)^n\right)^{\frac{1}{n}}<1+\left(\frac{b}{a}\right)^n$ , $\lim_{n\rightarrow\infty}\left(1+\left(\frac{b}{a}\right)^n\right)^{\frac{1}{n}}=1$ by squeeze theorem. ( $a\neq{b}$ )

If $a=b,\quad(a^n+b^n)^{\frac{1}{n}}=2^{\frac{1}{n}}\cdot{a}\rightarrow{a}$ when $n\rightarrow{\infty}$ .

So, the value we want to find is $a$ .

**Posted:** Sun Nov 01, 2009 6:58 am

Kent Merryfield

Some context.

**Posted:** Sun Nov 01, 2009 11:21 am

AndrewTom

Thanks, Kent.

**Posted:** Sun Nov 01, 2009 11:36 am