Step-by-step Solution:

To evaluate the limit as \( x \to \infty \), divide the numerator and the denominator by \( \sqrt{x} \).\n\n\[ L = \lim_{x \to \infty} \frac{\frac{\sqrt{x}}{\sqrt{x}}}{\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x}}} \]\n\[ = \lim_{x \to \infty} \frac{1}{\sqrt{\frac{x+\sqrt{x+\sqrt{x}}}{x}}} \]\n\[ = \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{\sqrt{x+\sqrt{x}}}{x}}} \]\n\nSimplify the term inside the square root further:\n\[ \frac{\sqrt{x+\sqrt{x}}}{x} = \sqrt{\frac{x+\sqrt{x}}{x^2}} = \sqrt{\frac{1}{x} + \frac{\sqrt{x}}{x^2}} = \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}}} \]\n\nAs \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( \frac{1}{x^{3/2}} \to 0 \).\nThus, \( \sqrt{\frac{1}{x} + \frac{1}{x^{3/2}}} \to 0 \).\n\nSubstitute this back into the limit:\n\[ L = \frac{1}{\sqrt{1 + 0}} = 1 \]