First slide

Introduction to limits

Question

$lim_{n \to \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n}, a, b > 0$ equals

Moderate

A

1
B

$\sqrt{a b}$
C

$a b$
D

$\frac{2 b}{2}$

Solution

$lim_{n \to \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n}$
$= lim_{n \to \infty} {(1 + \frac{a^{1 / n} + b^{1 / n} - 2}{2})}^{n}$
$= lim_{n \to \infty} {\{1 + \frac{(a^{1 / n} - 1) + (b^{1 / n} - 1)}{2}\}}^{n}$
$= e^{\lim_{n \to \infty} n \{\frac{(a^{1 / n} - 1) + (b^{1 / n} - 1)}{2}\}}$
$= e^{\frac{1}{2} lim_{n \to \infty} \{\frac{a^{1 / n} - 1}{1 / n} + {\frac{b^{1 / n}}{1 / n}}^{- 1}\}} = e^{1} (\log_{e} a + \log_{e} b) = e^{\log_{e} \sqrt{a b}} = \sqrt{a b} .$

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