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(5) \( \lim _{x \rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}} \)

Ask by Rojas Daniel. in Israel
Jan 23,2025

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Answer

\( \lim_{x\rightarrow\infty} \sqrt{x+\sqrt{x+\sqrt{x}}} = +\infty \)

Solution

Calculate or simplify the expression \( \lim_{x\rightarrow\infty} \sqrt{x+\sqrt{x+\sqrt{x}}} \). Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow +\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}\right)\) - step1: Rewrite the expression: \(\sqrt{\lim _{x\rightarrow +\infty}\left(x+\sqrt{x+\sqrt{x}}\right)}\) - step2: Calculate: \(\sqrt{+\infty}\) - step3: Simplify the root: \(+\infty\) אפשרות 5: \( \lim _{x \rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}} = +\infty \)

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The Deep Dive

To find the limit \( \lim_{x \rightarrow \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} \), we can start by simplifying the expression inside the square root. As \( x \) approaches infinity, we can notice that the terms \( \sqrt{x + \sqrt{x}} \) and \( \sqrt{x} \) become significant compared to any constant. Hence, we can analyze the dominant terms. 1. Begin with the inner square root: \[ \sqrt{x + \sqrt{x}} \approx \sqrt{x + x^{1/2}} \approx \sqrt{x} \text{ for large } x. \] This is because \( \sqrt{x} \) becomes negligible compared to \( x \). Specifically: \[ \sqrt{x + \sqrt{x}} = \sqrt{x(1 + \frac{1}{\sqrt{x}})} \approx \sqrt{x} \text{ as } x \to \infty. \] 2. Now, substitute this approximation into the outer square root: \[ \sqrt{x + \sqrt{x + \sqrt{x}}} \approx \sqrt{x + \sqrt{x}} \approx \sqrt{x} \text{ as } x \to \infty. \] 3. Therefore, we have: \[ \sqrt{x + \sqrt{x + \sqrt{x}}} \approx \sqrt{x}. \] 4. Finally, to find the limit: \[ \lim_{x \to \infty} \sqrt{x} = \infty. \] Thus, the limit is: \[ \boxed{\infty}. \]

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