Question-156677

Question Number 156677 by saly last updated on 14/Oct/21

Commented by cortano last updated on 14/Oct/21

\lim_{n \to \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n} ?

Commented by saly last updated on 14/Oct/21

w h a t h ?

Commented by john_santu last updated on 14/Oct/21

i f A = \lim_{x \to \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n} = {(\frac{\sqrt[n]{a} + \sqrt[b]{n}}{2})}^{n} \times \lim_{x \to \infty} (1)

= {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n}

Answered by puissant last updated on 14/Oct/21

A = \lim_{n \to + \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n}; x = \frac{1}{n} \to n = \frac{1}{x}

A = \lim_{x \to 0} {(\frac{a^{x} + b^{x}}{2})}^{\frac{1}{x}} = \lim_{x \to 0} {(\frac{1 + x l n a + 1 + x l n b}{2})}^{\frac{1}{x}}

= \lim_{x \to 0} {(1 + \frac{x}{2} l n (a b))}^{\frac{1}{x}} = \lim_{x \to 0} e^{\frac{l n (1 + \frac{x}{2} l n a b)}{x}}

= \lim_{x \to 0} e^{\frac{\frac{x}{2} l n a b}{x}} = e^{\frac{1}{2} l n a b} = \sqrt{a b} \dots

∴∵ A = \lim_{n \to + \infty} {(\frac{\sqrt[n]{a} + \sqrt[n]{b}}{2})}^{n} = \sqrt{a b} \dots .

\dots \dots \dots \dots L e p u i s s a n t \dots \dots \dots \dots

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