<p>Evaluate \(\lim_{x \to \infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}\right)\)</p>
Step-by-Step Solution
Key Concept: Factor out √x from the nested radical and use the conjugate method: multiply by (√(x+√(x+√x)) + √x)/(√(x+√(x+√x)) + √x) to convert the indeterminate form ∞-∞ into a determinate limit.
<p><strong>Step 1:</strong> Recognize ∞-∞ indeterminate form. Use conjugate multiplication:</p><p>$$\lim_{x \to \infty} \frac{(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x})(\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x})}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}$$</p><p><strong>Step 2:</strong> Simplify numerator:</p><p>$$= \lim_{x \to \infty} \frac{x + \sqrt{x + \sqrt{x}} - x}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} = \lim_{x \to \infty} \frac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}$$</p><p><strong>Step 3:</strong> Factor √x from nested radicals. For large x: √(x+√x) ≈ √x√(1+1/√x) ≈ √x and √(x+√(x+√x)) ≈ √x</p><p><strong>Step 4:</strong> Divide numerator and denominator by √x:</p><p>$$= \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{\sqrt{x}}}}{\sqrt{1 + \frac{\sqrt{x + \sqrt{x}}}{x}} + 1}$$</p><p><strong>Step 5:</strong> As x→∞, terms with 1/√x → 0 and √(x+√x)/x → 0:</p><p>$$= \frac{\sqrt{1 + 0}}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$</p><p>∴ Answer: <strong>1/2</strong></p>
Correct Answer: 1/2