Zeroed Nested root

Calculus Level 1

It's obvious that if d = 0 d = 0 , then d + d + d + d + \sqrt{d+\sqrt{d+\sqrt{d+\sqrt{d+\cdots}}}} is equal to 0.

But what is the value of the limit lim d 0 + d + d + d + d + ? \lim_{d\to0^+} \sqrt{d+\sqrt{d+\sqrt{d+\sqrt{d+\cdots}}}}\ ?


The answer is 1.

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2 solutions

Zico Quintina
May 10, 2018

α = d + d + d + d + α 2 = d + d + d + d + α 2 d = d + d + d + d + = α \begin{aligned} \alpha &= \sqrt{d + \sqrt{d + \sqrt{d + \sqrt{d + \ldots}}}} \\ \alpha^2 &= d + \sqrt{d + \sqrt{d + \sqrt{d + \ldots}}} \\ \alpha^2 - d &= \sqrt{d + \sqrt{d + \sqrt{d + \sqrt{d + \ldots}}}} \\ \\ \\ &= \alpha \\ \\ \end{aligned}

Rearranging,

α 2 α d = 0 α = 1 ± 1 + 4 d 2 \begin{aligned} \\ \alpha^2 - \alpha - d &= 0 \\ \\ \alpha &= \dfrac{1 \pm \sqrt{1 + 4d} }{2} \\ \\ \end{aligned}

We can discard the negative root as our nested radical is clearly positive. Then,

lim d 0 + d + d + d + d + = lim d 0 + 1 + 1 + 4 d 2 = 1 + 1 2 = 1 \\ \displaystyle \lim_{d \to 0^+} \sqrt{d + \sqrt{d + \sqrt{d + \sqrt{d + \ldots}}}} = \displaystyle \lim_{d \to 0^+} \dfrac{1 + \sqrt{1 + 4d} }{2} = \dfrac{1 + \sqrt{1} }{2} = \boxed{1}

Exponent Bot
May 19, 2018

We change d d to x x . We let y y denote the radical expression and graph the following s q r t ( x + y ) = y sqrt(x+y)=y . As x approaches 0 from the positive side, the expression approaches 1.

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