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

To evaluate the limit \[ \lim_{x \rightarrow \infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} \right) \] we start by expressing the term under the limit in a more manageable form. The phrase can be rewritten as: \[ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} = \frac{(\sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x})(\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x})}{\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}} \] However, it’s easier to focus directly on the limit itself. First, let's simplify \( \sqrt{x+\sqrt{x+\sqrt{x}}} \) as \( x \to \infty \). 1. Note that \( \sqrt{x+\sqrt{x+\sqrt{x}}} \) can be approximated: \[ \sqrt{x+\sqrt{x+\sqrt{x}}} \approx \sqrt{x+\sqrt{x}} = \sqrt{x + x^{1/2}} = \sqrt{x(1+x^{-1/2})} \] As \( x \to \infty \), \( x^{-1/2} \) approaches 0, so we approximate: \[ \sqrt{1+x^{-1/2}} \approx 1 + \frac{1}{2} x^{-1/2} \] Thus, \[ \sqrt{x+\sqrt{x+\sqrt{x}}} \approx \sqrt{x} (1 + \frac{1}{2} x^{-1/2}) = \sqrt{x} + \frac{1}{2} \] So, \[ \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} \approx \frac{1}{2} \] Consequently, we find: \[ \lim_{x \to \infty} \left( \sqrt{x+\sqrt{x+\sqrt{x}}} - \sqrt{x} \right) = \frac{1}{2} \] Therefore, the limit evaluates to: \[ \boxed{\frac{1}{2}} \]