What is \(\displaystyle\lim_{x\to\infty} \left(\sqrt{x+\sqrt{x}} - \sqrt{x}\right)\) equal to?
\(\dfrac{1}{2}\)
Writing \(\sqrt{x+\sqrt{x}} - \sqrt{x} = \sqrt{x}\left[\sqrt{1+\tfrac{1}{\sqrt{x}}} - 1\right]\) and using the expansion \(\sqrt{1+t} \approx 1 + \tfrac{t}{2}\) for small \(t = \tfrac{1}{\sqrt{x}}\), this becomes approximately \(\sqrt{x}\cdot\tfrac{1}{2\sqrt{x}} = \tfrac{1}{2}\) as \(x \to \infty\). Hence the limit equals \(\dfrac{1}{2}\).
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