Infinite Radicals

Calculus Level 2

$\begin{aligned} f(1) & =\sqrt{1} \\ f(2) & = \sqrt{\sqrt{2}} \\ \cdots & = \cdots \\ f(n) & = \underbrace{\sqrt{\sqrt{\sqrt{{\cdots \sqrt{n}}}}}}_{\text{Number of }\sqrt \ = n} \end{aligned}$

For $f(n)$ as defined above, what is $\displaystyle \lim_{n \to \infty} f(n)$ ?

e Goes to infinity 1 2

\pi

1 solution

Chew-Seong Cheong
Jul 17, 2018

$\begin{aligned} L &= \lim_{n \to \infty} n^{\frac 1{2^n}} \\ &= \lim_{n \to \infty} \exp \left(\ln n^{\frac 1{2^n}} \right) & \small \color{#3D99F6} \text{where }\exp(x) = e^x \\ &= \exp \left(\lim_{n \to \infty} \frac {\ln n}{2^n} \right) & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case L'Hôpital's rule applies.} \\ &= \exp \left(\lim_{n \to \infty} \frac {\frac 1n}{2^n\ln 2} \right) & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ &= e^0 = \boxed 1 \end{aligned}$

Thank you for adding this solution! This is my first ever problem so thank you for adding an explanation to it.

Jacob Moore - 2 years, 10 months ago

You are welcome.

Chew-Seong Cheong - 2 years, 10 months ago

Thank you! Feel free to check out my latest problem to make sure it and the solution are both correct

Jacob Moore - 2 years, 10 months ago

I have edited your problem too.

Chew-Seong Cheong - 2 years, 10 months ago

Infinite Radicals

1 solution

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