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Expert-verified Found in: Page 435 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Find the limits of the sequences$$\left( {\sqrt 2 ,\sqrt {2\sqrt 2 } ,\sqrt {2\sqrt {2\sqrt 2 } } ,........} \right)$$.

In the given question we will first write a general term and then find limits (tending to infinity)

See the step by step solution

## Given data:-

The sequence=$$\sqrt 2 .\sqrt {2\sqrt 2 } .\sqrt {2\sqrt {2\sqrt 2 } }$$

## Computing general terms:

\begin{aligned}{}{a_1} &= \sqrt 2 = {2^{1/2}}\\{a_2} &= \sqrt {2\sqrt 2 } = {\left( {{{2.2}^{1/2}}} \right)^{1/2}}= {2^{1/2}}{.2^{1/4}}= {2^{3/4}} \\ {a_3} &= \sqrt {2\sqrt {2\sqrt 2 } } = {\left( {{{2.2}^{1/2}}.{{\left( {{2^{1/2}}} \right)}^{1/2}}} \right)^{1/2}} = {2^{7/8}}\end{aligned}

Thus, we can say that

$${a_n} = {2^{\left( {\frac{{{2^n} - 1}}{{{2^n}}}} \right)}}$$

## Substituting limit $$n \to \infty$$

\begin{aligned}{}{a_n} &= \mathop {\lim }\limits_{n \to \infty } ({2^{\left( {\frac{{{2^n} - 1}}{{{2^n}}}} \right)}}) &= \mathop {\lim }\limits_{n \to \infty } \left( {{2^{\frac{{{2^n}}}{{{2^n}}} - \frac{1}{{{2^n}}}}}} \right)\\\end{aligned}

\begin{aligned}{}{a_n} &= \mathop {\lim }\limits_{n \to \infty } \left( {{2^{1 - \frac{1}{{{2^n}}}}}} \right) &= 2\\\end{aligned}

Hence, the required answer is 2. ### Want to see more solutions like these? 