4

$$\lim_{n\rightarrow\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n$$

No idea how to do it please help step by step

Martin Sleziak
  • 53,687
Kran
  • 1,039

2 Answers2

3

the natural log of both sides should yield an answer (after applying l'Hopitals rule).

\begin{align*} \lim_{n \to \infty}f(n)&= \left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n \\ \lim_{n \to \infty}\log n&= \frac{\log \left(\frac{a^{1/n}+b^{1/n}}{2}\right)}{1/n} \\ &=\lim_{n \to \infty}\frac{-(a^{1/n}\log(a) +b^{1/n}\log(b))/(2n^2)}{-1/n^2}\\ &=\lim_{n \to \infty}\frac{1}{2}\left(a^{1/n}\log(a) +b^{1/n}\log(b)\right)\\ &=\frac{1}{2}\log(ab) \end{align*} So $$\lim_{n \to \infty}f(n)=(ab)^{1/2}$$

Martin Sleziak
  • 53,687
recmath
  • 2,738
  • Wolfram alpha says otherwise http://www.wolframalpha.com/input/?i=limit+%28a%5E%281%2Fn%29%2Bb%5E%281%2Fn%29%29%5En%2F2%5En+n+to+infinity – Haha Nov 13 '13 at 01:05
  • You're right! sorry I made a grave error in using L'hipitals rule on the third line – recmath Nov 13 '13 at 01:07
  • on second thought I used L'hipital's right...it was the calculation in the fourth line that I messed up. I think this is the correct solution now. – recmath Nov 13 '13 at 01:20
  • Correct, but right it more carefully – Haha Nov 13 '13 at 01:23
1

Assume that there is a $y$ so that $$ \lim_{n\to\infty}\left(1+\frac{y}{n}\right)^n=x\tag{1} $$ Then $$ \lim_{n\to\infty}\left(\frac{1+\frac{y}{n}}{x^{1/n}}\right)^n=1\tag{2} $$ Since $\frac{1+a}{1+b}$ is between $\frac11$ and $\frac{a}{b}$, $(2)$ implies that $$ \begin{align} 1 &=\lim_{n\to\infty}\left(\frac{1+1+\frac{y}{n}}{1+x^{1/n}}\right)^n\\[6pt] &=\lim_{n\to\infty}\left(\frac{1+\frac{y}{2n}}{\frac{1+x^{1/n}}{2}}\right)^n\tag{3} \end{align} $$ Squaring $(3)$ and applying $(1)$ gives $$ \begin{align} x&=\lim_{n\to\infty}\left(1+\frac{y}{2n}\right)^{2n}\\[6pt] &=\lim_{n\to\infty}\left(\frac{1+x^{1/n}}{2}\right)^{2n}\tag{4} \end{align} $$ Taking the square root of $(4)$ yields $$ \lim_{n\to\infty}\left(\frac{1+x^{1/n}}{2}\right)^n=\sqrt{x}\tag{5} $$ Applying $(5)$ to the problem says $$ \begin{align} \lim_{n\to\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n &=a\lim_{n\to\infty}\left(\frac{1+(b/a)^{1/n}}{2}\right)^n\\ &=a\sqrt{b/a}\\[9pt] &=\sqrt{ab}\tag{6} \end{align} $$

robjohn
  • 345,667